Download mass mass calc

SCH 3U
Mole Ratios in Balanced Chemical Equations
Name: _________________
Date: _________________
Some New Definitions
1. Stoichiometry is the study of the quantitative relationships among the amounts of reactants used and
the amounts of products formed in a chemical reaction.
2. Mole ratios (pg. 212) express the ratio of the amounts (in moles) of any two substances in a balanced
chemical equation.
Practice:
1. Consider the following chemical equation when answering the questions below.
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
a) Write the mole ratio for ethane, C2H6(g) and carbon dioxide.
b) Write the mole ratio for ethane and oxygen gas.
c) Write the mole ratio for carbon dioxide gas and water.
2. More Practice: Do #3 page 305.
Using Mole Ratios from a Balanced Chemical Equation in a Stoichiometry Problem
Sample Problem:
How many moles of silver chromate, Ag2CrO4(s) is produced when 0.05 mol of silver nitrate, AgNO3(aq) reacts
according to the following balanced chemical equation?
2AgNO3(aq) + Na2CrO4(aq)  Ag2CrO4(s) + NaNO3(aq)
Solution:
You know from the balanced equation that 2 moles of AgNO3(aq) will produce 1 mol of Ag2CrO4(s). So, you can
set up a ratio….
2 mol AgNO3
0.05 mol AgNO3

1 mol Ag 2 CrO4
n Ag2CrO4
n Ag2CrO4  0.05 mol AgNO3 
1mol Ag 2 CrO4
2 mol AgNO3
= 0.025 mol of Ag2CrO4(s)
 If 0.05 mol of AgNO3 is used, 0.025 mol of Ag2CrO4(s) is produced.
If you are ratio impaired, try skipping right to this equation (it works every time):
nunknown = n known x ratio from the balanced equation arranged so that the appropriate units will cancel…
n

nunknown  nknown   unknown in balanced equation 
 nknown in balanced equation 


In this case:
n Ag2CrO4  n AgNO3 
n Ag2CrO4 in balanced equation
n AgNO3 in balanced equation
n Ag2CrO4  0.05 mol AgNO3 
1mol Ag 2 CrO4
2 mol AgNO3
= 0.025 mol of Ag2CrO4(s)
 If 0.05 mol of AgNO3 is used, 0.025 mol of Ag2CrO4(s) is produced.
Practice
Do the following questions. Use either of the methods above. All standard format rules apply.
1. Calculate the number of moles of water that will form when 6.00 moles of carbon dioxide is used in the
following reaction. 2NH3(g) + CO2(g)  NH2CONH2(s) + H2O(g)
2. Calculate the number of moles of oxygen that is needed to react with 2.40 moles of ammonia to produce
poisonous hydrogen cyanide, HCN(g) given the balanced chemical equation::
2NH3(g) + 3O2(g) + 2CH4(g)  2HCN(g) + 6H2O(g)
3. Calculate the number of moles of fluorine gas which will be needed to produce 2.35 mol of xenon
tetrafluoride. Xe(g) + 2F2(g)  XeF4(s)
4. Do #17, 19 page 332.
Mass-Mass Calculations (p. 301 - 305)
For this type of problem:
•
You will be given the mass of one substance in a chemical reaction and asked to calculate the mass of
another. For most of you, when working with balanced equations it will be best to work in moles rather than
in grams or number of particles so you will need to convert the given information from grams to moles to
begin the math.
•
Always start the solution with a balanced chemical equation. Place the known and unknown masses labeled
clearly underneath corresponding formulas.
•
In general, most of you will follow this pattern to solve for an unknown mass:
mass of the
required
substance
given mass
of some
substance
step 1
moles of
the given
substance
step 3
step 2
moles of the
required
substance
Mass-to-Mass Sample Problem Solved in Three Different Ways:
Given a mass of 10.2 g of C2H6, what mass of CO2 would be produced according to the balanced
chemical equation, 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) ?
Solution #1:
2 C2H6(g)
+ 7 O2(g)
Steps:
1. Given & Required Masses
10.2 g
[write them under the appropriate formula]
2. Molar masses of relevant
30.08
 4 CO2(g) + 6 H2O(g)
unknown
g
mol
44.01
g
mol
substances
3. Calculate the # moles of
the "given" substance
m
mm
10.2 g
=
g
30.08
mol
nC2H6 =
= 0.3390957 mol
4. Solve for the unknown number of moles using the mole ratios from the balanced equation.
nunknown = nknown x ratio from the balanced equation arranged so that the
appropriate units will cancel
= nknown x nunknown in balanced chem. equation
nknown in balanced chem. equation
4 mol CO2(g)
nCO2 = 0.339057 mol C2H6 x
2 mol C2H6(g)
nCO2 = 0.6781914 mol CO2
[Check that the answer makes sense...The balanced chemical equation shows that for every
2 moles of C2H6 used up, we should produce twice as much or 4 moles of CO2. In this experimental set up,
we started with 0.339057 mol of C2H6 and so should expect twice as many moles of CO2 at the end of the
reaction...
ie. we should expect 2 x 0.339057 mol = 0.6781914 mol of CO 2]
5. Now that the number of moles of CO2 produced is known, you can calculate the mass of CO2
produced in grams.
mCO2 = nCO2 x mmCO2
mass CO2 produced = 0.6781914 mol CO2 x
44.01 g
1 mol
= 29.847204 g
The mass of CO2 produced by 10.2 g of C2H6 is about 29.8 g.
Solution #2:
2C2H6(g) +
Steps:
1. Given & Required Masses
10.2 g
2. Molar masses of relevant
30.08
7O2(g)
 4CO2(g)
+ 6H2O(g)
unknown
g
mol
44.01
g
mol
substances
3. Relevant mole information
from balanced equation
2 mol C2H6(g)
2 mol x 30.08
g
= 60.16 g
mol
4 mol CO2(g)
4 mol x 44.01
g
= 176.04 g
mol
made into information
about mass
4. Solve for the unknown mass using the ratio established by the balanced chemical equation.
mass of CO2(g) = given mass of C2H6(g) x conversion factor using masses
derived from the balanced equation - the mass of C2H6(g) has
to be in the denominator in order for the mass of CO2 to be left.
176.04 g CO2
mCO2 = 10.2 g C2H6 x
60.16 g C2H6
= 29.847207 g of CO2
The mass of CO2 produced by 10.2 g of C2H6 is about 29.8 g.
Solution #3:
2C2H6(g) + 7O2(g) 
Steps:
1. Given & Required Masses
10.2 g
2. Molar masses of relevant
30.08
4CO2(g) + 6H2O(g)
unknown
g
mol
44.01
g
mol
substances
3. Write out and solve one equation using a variety of conversion factors to get from one unit to the next.
1mol C2H6
4 mol CO2
44.01 g CO2
Mole ratio from the
mCO2 = 10.2 g C2H6 x
x
x
30.08 g C2H6
2 mol C2H6
1mol CO2
balanced chemical
= 29.847207 g CO2
The mass of CO2 produced by 10.2 g of C2H6 is about 29.8 g.
equation written so that
the appropriate units
cancel.
Mass-Mass Problems: Solution Summary
Solution #1 shows the pattern most often used by students in this course because it tends to be the one
that is the most flexible when new problems are presented later on. Choose one of the three methods that
suits you best or make up your own. Regardless of how you choose to solve these problems, be sure to include
the following items in your work:
1. A copy of the balanced chemical equation must start the solution.
2. The known and unknown quantities should appear under the appropriate spot in the balanced equation.
3. The molar masses of relevant substances must be shown.
4. Clear step-by-step work must be shown.
5. Complete units must appear throughout the solution.
6. Round off only once, at the end of the solution, so that the final answer, presented in the summary
statement at the end of the solution, is the only one shown rounded off to the correct number of
significant digits.
Practice

Define the terms mole ratio (pg. 299) and stoichiometry”. (pg. 296).

Answer #2 page 305.

Do #4, 5, 6 page 305.

Do the problems below.
SCH3U STOICHIOMETRY PROBLEMS
1.
a)
b)
c)
Given the chemical equation:
3N2H4(l)  4NH3(g) + N2(g)
How many moles of N2(g) will be produced if 6.00 moles of N2H4(l) react completely?
(Ans 2.00 moles)
How many moles of NH3(g) will be produced if 6.00 moles of N2H4(l) react completely?
(Ans 8.00 moles)
How many moles of NH3(g) will be produced if 2.00 moles of N2H4(l) react completely?
(Ans 2.67 moles)
2.
In the reaction, Fe(s) + S(s)  FeS(s), what mass of iron is needed to react completely with
32.0 g of sulfur?
(Ans. 55.7 g)
4.
What mass of sulfurous acid can be produced when 128 g of sulfur dioxide completely reacts with water in
the following reaction,
SO2 + H2O  H2SO3.
(Ans. 164 g)
5.
When solid aluminum metal is heated in oxygen gas, solid aluminum oxide is formed.
a) Write a balanced chemical equation for this reaction.
b) Calculate the mass of aluminum oxide that can be obtained when
25.0 g of aluminum metal completely reacts.
(Ans. 47.2 g)
6.
An aqueous solution of ammonium hydroxide, NH4OH(aq), reacts with an aqueous solution of copper (II)
nitrate, Cu(NO3)2(aq) in a double displacement reaction.
a) Write the balanced chemical equation for this reaction.
b) Calculate the mass of ammonium hydroxide needed to react completely with 75.0 g of copper (II)
nitrate.
(Ans. 28.0 g)
7.
Calculate the mass of aluminum metal needed to replace all of the iron from 27.8 g of iron (III) oxide. The
balanced equation for this single displacement reaction is:
Al(s) + Fe2O3(aq)  Al2O3(aq) + Fe(s)
(Ans. 9.39 g)
8.
Given the following balanced chemical equation:
C(s) + O2(g)  CO2(g)
In a particular lab set up, 50.0 g of oxygen gas are available for the combustion of 25.0 g of carbon..
a) Calculate the number of moles of oxygen gas and carbon solid that are each available to react.
b) Calculate the number of moles of oxygen gas that will actually be needed to react with all of
the carbon available. (So, use the number of moles of carbon available as the “known” and the
number of moles of oxygen gas needed to react with it as the “unknown”.)
c) Comment on what the calculation in part b) indicates. One of the reactants is said to be in
excess and the other is said to be a “limiting reagent”. Which one is which? How do you know?
(Ans. Carbon in excess by 0.500 mol, oxygen gas is the limiting reagent. This means the reaction will stop
when all of the oxygen gas is used up, leaving 0.500 mol of C(s) unreacted.)
Limiting Reagents, Excess Reagents, % Yield: Introduction
Some Definitions
Term
Limiting reagent (pg. 306)
Excess reagent (pg. 306)
Theoretical Yield
(pg. 314)
Actual yield
 Define and give reasons
why actual yield may
differ from theoretical
yield. (pg. 314)
Percent yield or percentage
yield (% yield)
 Define and give the
equation. (pg. 316)
Factors that Affect the Actual
Yield of a Product
(Table 7.1, pg. 316)
Definiton