# Stoichiometry - DOC

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```					Limiting Reactants and Percent Yield      Name: _____________________________                           Page 1
Limiting reactant: the reactant that is completely consumed in the reaction.
   The limiting reactant is not present in sufficient quantity to react with all other reactants.
   The reaction stops when the limiting reactant is completely consumed.
   Any remaining reactants are considered "excess reactants".
   The amount of product formed is determined by the "limiting reactant".

Steps in solving a limiting reactant problem. When you read a problem and it has two
givens- You will either solve using limiting reagent instructions or if Percent is
mentioned you will use the formula for % Yield.

Limiting reagent
1. Write a balanced equation for the reaction.
2. Write the givens above the equation.
3. Pick a product (any Product) and Determine the moles (or grams) of product that
could be formed by each reactant.
3. The least amount in step #3 identifies the limiting reactant (write limiting). The
largest is the excess chemical (write excess).
4. (If needed) Use the limiting reagent to find the number of moles or grams of
product to determine the mass produced.
EXTRA: If the problem asks how much of the excess chemical is left, do a third
stoichiometry to determine the amount of excess used and subtract this number from
the amount of excess given in the problem.
Remember the following things about limiting reactants:
The reaction will stop when the reactants are used up.
If one reactant is used up before the other, the reaction stops then.
The first reactant used up is the limiting reactant, use it for the calculation.
The other reactant is the excess reactant, it is unimportant in the calculation.

An example of a limiting reactant problem:

What mass of water can be produced by 4 grams of H2 reacting with 16 grams of O2?

See the problem solution using the steps above and the next assignment.

The limiting reactant problem example:
What mass of water can be produced by 4g of H2 reacting with 16g of O2?

The problem solution:
1. Write a balanced equation for the reaction.

2 H2 + O2     2 H2O

2. Convert both reactant quantities to moles.
Limiting Reactants and Percent Yield         Name: _____________________________                 Page 2

3. Using the mole ratio from the equation, determine the moles (or grams) of water that
could be formed by each reactant.

4. Oxygen produces the least amount of water.
 16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other
words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen.
 Oxygen is the "limiting" reactant.
 Use oxygen for the calculation of product amount.

5. Complete the problem by converting moles of H2O to mass of H2O.

The theoretical yield for this problem is 18 grams. If you performed this reaction in the lab, your
actual yield might be less. Can you think of reasons why?

Book Work: Page 291 Practice problems 1-2, Page 294 Practice Problems 1-2
Section Review 2 -4, Page 297 25, 26, 27

Answers: Page 294 section review 2. b. 0.333 mol c. 19.4 g & 0.673 g/ 3. b. 15.9
g/ 4/ 93.7%/ Page 297 25. d. 468 g & 539 g/ 26. c. 34.9 g & 11.0 g/ 27. b. 2.35 
103 kg/ 29. 93.2 g/ 30. 1.02  104 g
Limiting Reactants and Percent Yield            Name: _____________________________              Page 3
1. A model airplane kit is designed to contain two wings, one fuselage, four engines, and six wheels. How
many model airplane kits can a manufacturer produce from a parts inventory of 426 wings, 224
fuselages, 860 engines, and 1578 wheels? (ans. 213 kits)

2. At high temperatures and pressures, nitrogen will react with hydrogen to produce ammonia as shown for
each of the following combinations of reactants, decide which is the limiting reactant. (ans. a) 3.65
moles of H2, b) 3.00 moles of H2, c) 55.0 g of N2)
a) 1.25 moles of N2 and 3.65 moles of H2

b) 44.0 g of N2 and 3.00 moles of H2

c) 55.0 g of N2 and 15.0 g of H2
Limiting Reactants and Percent Yield         Name: _____________________________               Page 4
3. Determine the number of grams of each of the products that can be made from 8.00 g of SCl2 and 4.00 g
of NaF by the following reaction.
3 SCl2 + 4 NaF --> SF4 + S2Cl2 + 4 NaCl (ans. 2.57 g SF4, 3.21 g S2Cl2, 5.57 g NaCl)

4. For the reaction: potassium + oxygen potassium oxide (ans. O2 will be the limiting reagent and
0.59 g K2O will be formed.)
a. Is 0.65 moles of O2 enough to react with 0.56 moles of K? (yes)

b. Is 0.65 g of O2 enough to react with 0.56 g of K? (yes)

c. How many grams of K2O will be produced from 0.50 g of K and 0.10 g of O2? (.59 g K2O Formed)
Limiting Reactants and Percent Yield      Name: _____________________________            Page 5
5. For the reaction: sodium oxide + water sodium hydroxide (ans. Na2O will be the limiting reagent
and 16.0 g NaOH will be formed.)
d. What weight of NaOH could be made from 12.4 g of Na2O and 42.1 g of H2O? (16.0 g NaOH)

e. What would be the limiting reagent if 100 g each of Na2O and H2O were allowed to react? (Na2O
limiting)

6. For the reaction: carbon + Hydrogen CH4 (ans. C will be the limiting reagent and 6.7 g CH4 will
be formed.)
f. How many moles of CH4 can be made from 7.0 moles of H2 and 5.0 moles of C? (3.5 moles CH4)

g. What weight of CH4 will be made when 10.0 g of H2 reacts with 5.0 g of C? (6.7 g CH4)
Limiting Reactants and Percent Yield         Name: _____________________________                Page 6
Work these limiting reactant practice problems: Answers below
7. Seventy five grams (exactly) of calcium oxide react with one hundred thirty grams (exactly) of
hydrochloric acid to produce a salt and water. What is the limiting reactant?

8. Five grams (exactly) of copper metal react with a solution containing twenty grams of silver nitrate to
produce copper (II) nitrate and silver.
a. What is the limiting reactant?

b. How much of the limiting reactant would be needed to react completely with the given amount of
excess reactant?

9. How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?
Limiting Reactants and Percent Yield           Name: _____________________________                Page 7
10. · (All Do This Problem) A solution containing 20.0 g of sodium sulfite reacts with 7.0 ml of phosphoric
acid. The concentration of the acid solution is such that there are 1.83 grams of H3PO4 per milliliter of
solution. Determine the following:
a. The mass of the excess reactant remaining at completion.

b. Grams of water produced.

c. Moles of sodium phosphate produced.

d. Grams of sulfur dioxide produced.

7 - calcium oxide                      9. 79 g of aluminum oxide               10c - 0.106 mole Na3PO4
8a - AgNO3                             10a -2.51 g H3PO4                       10d - 10.15 g SO2
8b - 26.6 g AgNO3                      10b - 2.85 g H2O
Stoichiometry                                                                                    Page 8 of 15
11. Barium sulfate is used in the "barium cocktail" given to patients prior to x-raying their intestinal tracts.
Based on the following equation, a chemist began with 75.00 grams of barium nitrate and enough sodium
sulfate to complete the reaction. After collecting and drying the product, 64.45 grams of barium sulfate
was obtained. What is the percentage yield of the reaction? (Hint: find the grams barium sulfate expected
using stoichiometry and compare using the formula on the cheat board.)

Theoretical and Percent Yield:
1. If 74.30 g of HCl were produced from 2.130 g of hydorgen and an excess of chlorine according to the
reaction what was the percent yield of hydrogen chloride? (ans. 96.44 %)

2. Aluminum and sulfur react to form aluminum sulfide by the equation. In a certain experiment, 125 g of
aluminum sulfide are produced from 75.0 g of Al and 300.0 g of Sulfur. (ans. a) 208 g Al2S3, b) 60.1 %
Al2S3)
a) What is the theoretical yield of Al2S3 ?

b) What is the percent yield of Al2S3 ?
Stoichiometry                                                                                Page 9 of 15
3. If the percent yield for the reaction 2 CO + O2  2 CO2 were 57.8%, what mass of product, in grams,
could be produced from a reactant mixture containing 35.0 g of each reactant? (ans. 31.8 g CO2)

Percent Yield: Although we can write perfectly balanced equations to represent perfect reactions, the
reactions themselves are often not perfect. A reaction does not always produce the quantity of products that
the balanced equation seems to guarantee. This happens not because the equation is wrong but because
reactions in the real world seldom produce perfect results. As an example of an imperfect reaction, look
again at the equation that shows the industrial production of ammonia.
N2(g) _ 3H2(g) : 2NH3(g)
In the manufacture of ammonia, it is nearly impossible to produce 2 mol (34.08 g) of NH3 from the simple
reaction of 1 mol (28.02 g) of N2 and 3 mol (6.06 g) of H2 because some ammonia molecules begin breaking
down into N2 and H2 molecules as soon as they are formed.
There are several reasons that real-world reactions do not produce products at a yield of 100%. Some are
simple mechanical reasons, such as:
• Reactants or products leak out, especially when they are gases.
• The reactants are not 100% pure.
• Some product is lost when it is purified.
There are also many chemical reasons, including:
• The products decompose back into reactants (as with the ammonia process).
• The products react to form different substances.
• Some of the reactants react in ways other than the one shown in the equation. These are called side
reactions.
• The reaction occurs very slowly. This is especially true of reactions involving organic substances.
Chemists are very concerned with the yields of reactions because they must find ways to carry out reactions
economically and on a large scale. If the yield of a reaction is too small, the products may not be
competitive in the marketplace. If a reaction has only a 50% yield, it produces only 50% of the amount of
product that it theoretically should. In this chapter, you will learn how to solve problems involving real-
world reactions and percent yield.
Book work: Page 297 # 29, 30
Page 932 – 935
Problem # 204b, 207a, 210b, 213c, 214, 217b, 219a&b, 222b&c, 223a&c,
224c, 225a&b, 232, 233b, 236
204b-16 g                           217b- 2.3 to 2.5 g                        225a- 81 % & b-200 g
207a-40 g                           219a- 79 % & b 75+%                       232 90%
210b-1 g                            222b- 80% & c- 2 X105 mol                 233b 90 %
213c-150 kg                         223a- 90% & c- 16 g                       236- 28 kg
214- 30 g remain                    224a- 87 % & c- 30,000 &44,000 kg
Stoichiometry                  Name: _____________________________________                         Page 10 of 15
1. In the commercial production of the element arsenic, arsenic(III) oxide is heated with carbon, which
reduces the oxide to the metal (arsenic) and carbon dioxide.
a. If 8.87 g of As2O3 is used in the reaction and 5.33 g of As is produced, what is the percent yield?
b. If 67 g of carbon is used up in a different reaction and 425 g of As is produced, calculate the percent
yield of this reaction.
2. Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating agent in research. It can be
produced by passing chlorine gas over heated mercury(II) oxide according to the following equation:
Mercury II + dichlorine monobromide  mercury II chloride + dichlorine monoxide
What is the percent yield, if the quantity of reactants is sufficient to produce 0.86 g of Cl2O but only 0.71
g is obtained?
3. Acetylene, C2H2 , can be used as an industrial starting material for the production of many organic
compounds. Sometimes, it is first brominated to form 1,1,2,2-tetrabromoethane, CHBr2CHBr2 , which
can then be reacted in many different ways to make other substances. The equation for the bromination
of acetylene follows:

If 72.0 g of C2H2 reacts with excess bromine and 729 g of the product is recovered, what is the percent
yield of the reaction?
4. Assume the following hypothetical reaction takes place.
2A + 7B  4C + 3D
Calculate the percent yield in each of the following cases:
a. The reaction of 0.0251 mol of A produces 0.0349 mol of C.
b. The reaction of 1.19 mol of A produces 1.41 mol of D.
c. The reaction of 189 mol of B produces 39 mol of D.
d. The reaction of 3500 mol of B produces 1700 mol of C.
5.    The reaction Ethyl acetate is a sweet-smelling solvent used in varnishes and fingernail-polish remover.
It is produced industrially by heating acetic acid and ethanol together in the presence of sulfuric acid,
which is added to speed up the reaction. The ethyl acetate is distilled off as it is formed. The equation
for the process is as follows.

Determine the percent yield in the following cases:
a. 68.3 g of ethyl acetate should be produced but only 43.9 g is recovered.
b. 0.0419 mol of ethyl acetate is produced but 0.0722 mol is expected. (Hint: Percent yield can also be
calculated by dividing the actual yield in moles by the theoretical yield in moles.)
c. 4.29 mol of ethanol is reacted with excess acetic acid, but only 2.98 mol of ethyl acetate is produced.
d. A mixture of 0.58 mol ethanol and 0.82 mol acetic acid is reacted and 0.46 mol ethyl acetate is
produced. (Hint: What is the limiting reactant?)
Stoichiometry                Name: _____________________________________                Page 11 of 15
6. Elemental phosphorus can be produced by heating calcium phosphate from rocks with silica sand (SiO2)
and carbon in the form of coke. The following reaction takes place.
Calcium phosphate + silicon trioxide + carbon  Calcium silicate + phosporus + carbon monoxide
a. If 57 mol of Ca3(PO4)2 is used and 101 mol of CaSiO3 is obtained, what is the percent yield?
b. Determine the percent yield obtained if 1280 mol of carbon is consumed and 622 mol of CaSiO3 is
produced.
c. The engineer in charge of this process expects a yield of 81.5%. If 1.4 x 105 mol of Ca3(PO4)2 is
used, how many moles of phosphorus will be produced?
7.     Tungsten (W) can be produced from its oxide by reacting the oxide with hydrogen at a high temperature
according to the following unbalanced equation:
WO3 + hydrogen  tungsten + water
a. What is the percent yield if 56.9 g of WO3 yields 41.4 g of tungsten?
b. How many moles of tungsten will be produced from 3.72 g of WO3 if the yield is 92.0%?
c. A chemist carries out this reaction and obtains 11.4 g of tungsten. If the percent yield is 89.4%, what
mass of WO3 was used?
8.     Carbon tetrachloride, CCl4 , is a solvent that was once used in large quantities in dry cleaning. Because
it is a dense liquid that does not burn, it was also used in fire extinguishers. Unfortunately, its use was
discontinued because it was found to be a carcinogen. It was manufactured by the following unbalanced
reaction:
Carbon disulfide + Chlorine  carbon tetrachloride + disulfur dichloride
The reaction was economical because the byproduct disulfur dichloride, S2Cl2 , could be used by
industry in the manufacture of rubber products and other materials.
a. What is the percent yield of CCl4 if 719 kg is produced from the reaction of 410. kg of CS2 .
b. If 67.5 g of Cl2 are used in the reaction and 39.5 g of S2Cl2 is produced, what is the percent yield?
c. If the percent yield of the industrial process is 83.3%, how many kilograms of CS2 should be
reacted to obtain 5.00 x 104 kg of CCl4 ? How many kilograms of S2Cl2 will be produced, assuming
the same yield for that product?
9.     Nitrogen dioxide, NO2 , can be converted to dinitrogen pentoxide, N2O5 , by reacting it with ozone, O3 .
The reaction of NO2 takes place according to the following equation:
2NO2(g) + O3(g)  N2O5(s or g) + O2(g)
a. Calculate the percent yield for a reaction in which 0.38 g of NO2 reacts and 0.36 g of N2O5 is
recovered.
b. What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there is a 61.1% yield in the
reaction?
10. In the past, hydrogen chloride, HCl, was made using the salt-cake method. This reacts sodium chloride
with sulfuric acid. If 30.0 g of NaCl and 0.250 mol of H2SO4 are available, and 14.6 g of HCl is made,
what is the percent yield?
Stoichiometry                  Name: _____________________________________                     Page 12 of 15
11. Cyanide compounds such as sodium cyanide, NaCN, are especially useful in gold refining because they
will react with gold to form a stable compound that can then be separated and broken down to retrieve
the gold. Ore containing only small quantities of gold can be used in this form of “chemical mining.”
The equation for the reaction follows.
4Au + 8NaCN + 2H2O + O2  4NaAu(CN)2 + 4NaOH
a. What percent yield is obtained if 410 g of gold produces 540 g of NaAu(CN)2?
b. Assuming a 79.6% yield in the conversion of gold to NaAu(CN)2 , what mass of gold would
produce 1.00 kg of NaAu(CN)2 ?
c. Given the conditions in (b), what mass of gold ore that is 0.001% gold would be needed to produce
1.00 kg of NaAu(CN)2 ?
12. Diiodine pentoxide is useful in devices such as respirators because it reacts with the dangerous gas
carbon monoxide, CO, to produce relatively harmless CO2 according to the following unbalanced
equation:
Diiodine pentoxide + carbon monoxide  iodine + carbon dioxide
a. In testing a respirator, 2.00 g of carbon monoxide gas is passed through diiodine pentoxide. Upon
analyzing the results, it is found that 3.17 g of I2 was produced. Calculate the percent yield of the
reaction.
b. Assuming that the yield in (a) resulted because some of the CO did not react, calculate the mass of
CO that passed through.
13. Sodium hypochlorite, NaClO, the main ingredient in household bleach, is produced by bubbling
chlorine gas through a strong lye (sodium hydroxide, NaOH) solution. The following unbalanced
equation shows the reaction that occurs.
Sodium hydroxide (aq) + chlorine (g)  sodium chloride(aq) + sodium hypochlorite(aq) + water(l)
a. What is the percent yield of the reaction if 1.2 kg of Cl2 reacts to form 0.90 kg of NaClO?
b. If a plant operator wants to make 25 metric tons of NaClO per day at a yield of 91.8%, how many
metric tons of chlorine gas must be on hand each day?
c. What mass of NaCl is formed per mole of chlorine gas at a yield of 81.8%?
d. At what rate in kg per hour must NaOH be replenished if the reaction produces 370 kg/h of NaClO at
a yield of 79.5%? Assume that all of the NaOH reacts to produce this yield.
14. Magnesium burns in oxygen to form magnesium oxide. However, when magnesium burns in air, which
is only about 1/5 oxygen, side reactions form other products, such as magnesium nitride, Mg3N2 .
a. Write a balanced equation for the burning of magnesium in oxygen.
b. If enough magnesium burns in air to produce 2.04 g of magnesium oxide but only 1.79 g is obtained,
what is the percent yield?
c. Magnesium will react with pure nitrogen to form the nitride, Mg3N2 . Write a balanced equation for
this reaction.
d. If 0.097 mol of Mg react with nitrogen and 0.027 mol of Mg3N2 is produced, what is the percent yield
of the reaction?
Stoichiometry                Name: _____________________________________                        Page 13 of 15
15. Some alcohols can be converted to organic acids by using sodium dichromate and sulfuric acid. The
following equation shows the reaction of 1-propanol to propanoic acid.
3CH3CH2CH2OH + 2Na2Cr2O7 + 8H2SO4  3CH3CH2COOH + 2Cr2(SO4)3 + 2Na2SO4 + 11H2O
a. If 0.89 g of 1-propanol reacts and 0.88 g of propanoic acid is produced, what is the percent yield?
b. A chemist uses this reaction to obtain 1.50 mol of propanoic acid. The reaction consumes 136 g of
propanol. Calculate the percent yield.
c. Some 1-propanol of uncertain purity is used in the reaction. If 116 g of Na2Cr2O7 are consumed in the
reaction and 28.1 g of propanoic acid are produced, what is the percent yield?
16. Acrylonitrile, C3H3N(g), is an important ingredient in the production of various fibers and plastics.
Acrylonitrile is produced from the following reaction:
C3H6(g) + ammonia(g) + oxygen(g)  C3H3N(g) + water(g)
If 850. g of C3H6 is mixed with 300. g of NH3 and unlimited O2, to produce 850. g of acrylonitrile, what
is the percent yield? You must first balance the equation
17. Methanol, CH3OH, is frequently used in race cars as fuel. It is produced as the sole product of the
combination of carbon monoxide gas and hydrogen gas.
a. If 430. kg of hydrogen react, what mass of methanol could be produced?
b. If 3.12 x 103 kg of methanol are actually produced, what is the percent yield?
18. The compound, C6H16N2 , is one of the starting materials in the production of nylon. It can be prepared
from the following unbalanced reaction involving adipic acid, C6H10O4 :
Adipic acid(l) + ammonia(g) + hydorgen(g)  C6H16N2(l) + water
What is the percent yield if 750. g of adipic acid results in the production of 578 g of C6H16N2 ?
19. Plants convert carbon dioxide to oxygen during photosynthesis according to the following equation:
Carbon dioxide + water  glucose (C6H12O6) + oxygen
Balance this equation, and calculate how much oxygen would be produced if 1.37 x 104 g of carbon
dioxide reacts with a percent yield of 63.4%.
20. Lime, CaO, is frequently added to streams and lakes which have been polluted by acid rain. The
calcium oxide reacts with the water to form a base (hydroxide) that can neutralize the acid. If 2.67 x
102 mol of base are needed to neutralize the acid in a lake, and the above reaction has a percent yield of
54.3%, what is the mass, in kilograms, of lime that must be added to the lake?
Stoichiometry                Name: _____________________________________                       Page 14 of 15
10 Percentage Yield
1. ans: 80%, b. ans: 75%
2. 83% yield
3. 76.3% yield
4. a. 64.3% yield, b. 58.0% yield, c. 69.5% yield, d. CH3CH2OH is limiting; 79% yield
5. a. 69.5% yield, b. 79.0% yield, c. 48% yield, d. 85% yield
6. 3. a. 59% yield, b. 81.0% yield, c. 2.3 x 105 mol P
7. a. 91.8% yield, b. 0.0148 mol W, c. 16.1 g WO3
8. a. 86.8% yield, b. 92.2% yield, c. 2.97 x 104 kg CS2, 4.39 x 104 kg S2Cl2
9. a. 81% yield b. 2.0 x 102 g N2O5
10. 80.1% yield
11. a. 90-100% yield, b. 900 g Au, c. 90000 kg ore
12. a. 90% yield, b. ¼ g CO
13. a. 70% yield, b. 25-30 metric tons, c. 50 g NaCl, d. 500 kg per hour NaOH
14. a. equation, b. 90% yield, c. product is magnesium nitride, d. 55-60% yield
15. a. 80.% yield, b. 66-70% yield, c. 55-60% yield
16. 90% yield
17. a. equation 3.400 kg, b. 91.5% yield
18. 96.9% yield
19. 6.300 g O2
20. 25-30 kg CaO

Theoretical Yield
Purpose:
To determine the theoretical and experimental yields of NaCl. Also, calculate the
percentage yield.

Procedure:
1. Weigh a small beaker and a watch glass.
2. Weigh approximately 2 grams of NaHCO3 (baking soda) into the beaker. Record the exact mass. Use
the 0.01 g or 0.001 g balances.
3. Add 1 M HCl with an eye dropper until the reactions is complete. (No more bubbles) Then add a
little excess HCl.
4. Cover the beaker with the watch glass to keep the product from spattering on the counter. Evaporate
the solution to dryness. Be sure that all the moisture is gone from the beaker and watch glass.
5. While you wait for the evaporation to finish, calculate the theoretical yield of NaCl.
6. When completely dry, weigh the beaker, watch glass and product.
7. Wash all glass ware and put away.
8. Calculate the percentage yield of NaCl.
9. Turn in lab report: title, data table, analysis (all calculations), and conclusion.
Stoichiometry                    Name: _____________________________________                       Page 15 of 15
Does the problem
have more than 2
yes
no                                   chemicals in it?
Write a balanced
Does it have more                      equation.
than one given?
What are you looking for?
no                                yes
Mass?             Volume?          Moles?           Molarity?
Xg=               XL=              X mol=           XM=                                                                     Does it ask
Do standard 3 step                      for % yield?
stoichiometry.
Know?                   Know?      Know?                You are given a       X g = g 1mole #mole gmm
mass (or moles)               gmm #mole 1 mole              no                       yes
and a volume.
 Moles                                                   x M = g mol
Does it give                  Do 3 step
mol x gmm/mol                                                 L gmm
 Molecules
temperature or                stoichiometry
molec x gmm/avagadro’s         Molecules                                                                       pressure?                      and then
/#                              molec x mole/avagadro’s /#                                                                                   calculate %
 Volume of a gas                Mass                                              Molarity?                                                     yield.
L or mL x gmm/22.4L             g x 1 mole/ gmm                                                         no
 Volume of a solution           Volume of a gas                                                                                 yes
L x M (molarity mole/L)         L x mol/22.4
yes
 Volume of a solution M                                         no
given.                                                                                          Gas Law problem.
L x M (mol/L)                                                       Limiting                         PV= nRT
Is a volume given?
 Moles of a gas mol x 22.4L/mol                                                                     Reagent.                     Is Volume given?
 Molecules of a gas                                                                                 Pick a product
molec x 22.4 L/avagadro’s /#                                                                       find the sm. #
 Mass of a gas                                                no                      yes           moles do 3 step
stoichiometry
no                     yes
g x 22.4L/ gmm
 Temp &/or Pressure use Combined Gas
Law V2P2T1 = V1P1T2                            X vol = g mol mol           Xg=
 Temp, pressure, mass/mole use Ideal Gas                  gmm mol           vol M (mol) mol gmm     Given probably “g”                 Use PV/RT = n to
Law PV = nRT or PV = mRT/M                     then calculate the M             L      mol mol     xV = g mole mole nRT              solve for moles then
 Mass of a solution                             from ans.                                                   gmm mole P                PV mole gmm
g x gmm/mol x M (molarity flipped L/mol)
RT mole mole

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