Name:_________________________
Date:____________ Period:________
Another Way of Thinking About Stoichiometry
Use the diagram and/or steps below to set up and solve the problems underneath:
Mass of Given
Chemical
given chemical
mass
1 mole of given
chemical
molar mass of
given chemical
1
Mole Ratio
(use the chemical
equation)
Molar Mass of
Wanted
Material
moles of wanted
chemical (from
chemical equation)
molar mass of
wanted chemical
Molar Mass of
Given Chemical
moles of given
chemical (from
chemical equation)
1 mole of wanted
chemical
See Step 3
See Step 4
See Step 2
Mass of
Wanted
Material
To Solve:
multiply
everything
up here
divide
everything
down here
=
mass of
wanted
chemical
Written below is a sample problem that details the written steps that are necessary for solving a stoichiometry
problem. There is also a sample diagram showing how to take that same information and use it according to the
table above.
(adapted from http://www.jiskha.com/display.cgi?id=1157853894)
Sample Problem: You react 7.92 grams of silver nitrate, AgNO3(aq), with an excess of sodium chloride,
NaCl(aq). How many grams of silver chloride, AgCl(s), will precipitate from the solution?
Step 1. Write a balanced chemical equation.
ex. AgNO3 + NaCl ==> AgCl + NaNO3
Step 2. Convert what you are starting with or given to moles remembering that moles = grams/molar mass. The
molar mass is approximately 170 g. You will need to look up the exact value and use it.
ex. moles AgNO3 = 7.92 g AgNO3 / 170 g AgNO3= 0.0466
Step 3. Convert moles of what you have to moles of what you want using the coefficients in the chemical
equation from step 1.
ex. 0.0466 mole AgNO3 x (1 mole AgCl / 1 mole AgNO3) = 0.0466 mole AgCl.
*Note how the mol AgNO3 in the numerator cancels with the mols in the denominator to leave mols of
what you want; i.e., moles AgCl.
Step 4. Now convert moles of what you have into grams using the reverse of step 2.
moles = grams/molar mass, so grams = moles x molar mass.
grams AgCl = moles AgCl x molar mass AgCl
grams AgCl = 0.0466moles AgCl x 143 g AgCl = 6.66 grams
Here is an example of how to use the graphic organizer with the problem above:
Mass of Given
Chemical
7.92 g AgNO3
1
Molar Mass of
Given Chemical
1 mole AgNO3
170 g AgNO3
Mole Ratio (use the
chemical equation)
Molar Mass of
Wanted Chemical
1 mole AgCl
143 g AgCl
1 mole AgNO3
1 mole AgCl
Mass of Wanted
Chemical
=
6.66 g AgCl
Name:_________________________
Date:____________ Period:________
1. a) Chlorine gas (Cl2) reacts sodium (Na) to produce sodium chloride (NaCl). Balance the following equation.
_____ Na + _____ Cl2  _____ NaCl
b) Using the equation (after it is balanced) above, determine the amount of product that can be
produced from 24.7 g Na.
Mass of Given
Chemical
Molar Mass of
Given Chemical
_______ g ____ 1 mole ________
1
_______ g ____
Mole Ratio (use the
chemical equation)
Molar Mass of
Wanted Chemical
____ mole _____
_______ g ____
____ mole _____
1 mole _______
Mass of Wanted
Chemical
=
2. In a reaction, 10.0 grams pentane (C5H12) was burned to create water and carbon dioxide:
a) Balance the equation: _____ C5H12 + _____ O2  _____ CO2 + _____ H2O
b) How many grams of water were formed?
Mass of Given
Chemical
Molar Mass of
Given Chemical
_______ g ____ 1 mole ________
1
_______ g ____
Mole Ratio (use the
chemical equation)
Molar Mass of
Wanted Chemical
____ mole _____
_______ g ____
____ mole _____
1 mole _______
Mass of Wanted
Chemical
=
3. If ammonia, NH3, is burned in air, a reaction takes place:
a) Balance the equation: _____ NH3 + _____ O2  _____ N2 + _____ H2O
b) Given that you started with 51.0 g of NH3, how many g of water will be produced?
Mass of Given
Chemical
Molar Mass of
Given Chemical
Mole Ratio (use the
chemical equation)
Molar Mass of
Wanted Chemical
Mass of Wanted
Chemical
=
1
4. Consider the reaction below and balance it.
a) _____ I2O5(g) + _____ CO(g)  _____ CO2(g) + _____ I2(g)
b) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO.
Determine the mass of iodine I2, which could be produced?
=
1
=
1
Answers (in random order): 15.0 g 50.8 g 60.8 g 62.7 g 81.0 g