# Final Exam Review 2010 UbD

```Unit 6 Stoichiometry:
1. Define “limiting reactants” A limiting reactant is the reactant that runs out first in a chemical
reaction. It limits the amount of product you can make, because once it runs out the reaction will
stop.
How do you determine the limiting reactant for a chemical reaction? Use stoichiometry to determine
how much product each of your given reactants can produce. Whichever reactant produces the
LEAST amount of product is your limiting reactant because once it runs out the reaction must
stop!
2. Define “ percent yield” Percent Yield is the ratio of what was actually produced in a reaction (actual
yield) to the maximum possible amount of product (Theoretical Yield). It is then multiplied by 100
to make it a percentage. It should never exceed 100% as this would violate the law of
conservation of mass.
What is the formula for percent yield?
% Yield =
Actual Yield
Theoretical Yield
x 100
3. What do the coefficients in chemical equation represent? The coefficients represent the number of
molecules (or moles) of each substance in a chemical reaction.
4. Balance the reaction and use it to answer the rest of the questions.
6 H2O + 1 Mg3N2 → 3 Mg(OH)2 + 2 NH3
5. How many moles of Mg3N2 are needed to react with 15.2 moles of H2O?
Mol Ratio: 1 mol Mg3N2 : 6 mol H2O
15.2 moles of H2O
1 mol Mg3N2
6 mol H2O
=
2.53 mol Mg3N2
Steps to solve a Stoichiometry Problem
1. Use molar mass to convert given amount into moles (if it is already in moles skip this step)
2. Use the coefficients from the balanced equation to create mole ratio and convert from moles of what
was given to moles of what you want.
3. Use molar mass to convert moles of what you want into grams (if the answer is asked for in moles,
skip this step)
* This problem gave you a starting amount in moles and wanted the answer in moles, so the only step
you had to perform was #2 – Mole Ratio.
6 H2O + 1 Mg3N2 → 3 Mg(OH)2 + 2 NH3
6. If 10.5 grams of Mg3N2 are reacted, how many grams of Mg(OH)2 will be formed?
Mg = 24.31 x 3 = 72.93
N = 14.01 x 2 = + 28.02
1 mol Mg3N2 = 100.95 g Mg3N2
10.5 grams of Mg3N2
Mol Ratio:
1 mol Mg3N2 : 3 mol Mg(OH)2
1 mol Mg3N2
100.95 g Mg3N2
3 mol Mg(OH)2
1 mol Mg3N2
Mg = 24.31 x 1 = 24.31
O = 16.00 x 2 = 32.00
H = 1.01 x 2 = + 2.02
1 mol Mg(OH)2 = 58.33 g Mg(OH)2
58.33 g Mg(OH)2
1 mol Mg(OH)2
= 18.2 g Mg(OH)2
Steps to solve a Stoichiometry Problem
1. Use molar mass to convert given amount to moles (if it is already in moles skip this step)
2. Use the coefficients from the balanced equation to create mole ratio and convert from moles of what
was given to moles of what you want.
3. Use molar mass to convert moles of what you want to grams (if the answer is asked for in moles, skip
this step)
* This step gave you a starting amount in grams and wanted the answer in grams, so all 3 steps had to
be completed.
6 H2O + 1 Mg3N2 → 3 Mg(OH)2 + 2 NH3
7. In lab, 10.5 grams of Mg3N2 react with enough water to form 1.8 grams of NH3. What is the % yield of
NH3 in this lab?
Mg = 24.31 x 3 = 72.93
Mol Ratio:
N = 14.01 x 2 = + 28.02
1 mol Mg3N2 : 2 mol NH3
1 mol Mg3N2 = 100.95 g Mg3N2
10.5 grams of Mg3N2
1 mol Mg3N2
100.95 g Mg3N2
% Yield = Actual Yield
Theoretical Yield
x 100
%
2 mol NH3
1 mol Mg3N2
% Yield =
N = 14.01 x 1 = 14.01
H = 1.01 x 3 = + 3.03
1 mol NH3 =
17.04 g NH3
17.04 g NH3
1 mol NH3
=
3.54 g NH3
1.8 g NH3 . x 100 = 50.8 % Yield
3.54 g NH3
Steps to solve a % Yield Problem
1. Use Stoichiometry to determine the maximum amount of product you could have produced from the
reactant amount given. This is called your Theoretical Yield.
2. Locate the actual yield for one of the products given to you in the problem.
3. Use the % Yield formula to calculate the ratio of actual yield to theoretical yield as a percentage.
6 H2O + 1 Mg3N2 → 3 Mg(OH)2 + 2 NH3
8. If 15 moles of H20 and 4 moles of Mg3N2 are reacted together, how many moles of NH3 will be formed?
Which is the limiting reactant?
Mol Ratio:
6 moles H2O : 2 moles NH3
Mol Ratio:
1 mole Mg3N2 : 2 moles NH3
15 moles of H2O
2 moles NH3
6 moles H2O
= 5.0. moles NH3
4 moles of Mg3N2
2 moles NH3
1 mole Mg3N2
= 8 moles NH3
(This amount cannot be reached!)
If 15 moles of H2O and 4 moles of Mg3N2 react, you can expect to produce: