
How to convert on mass to another mass
o Typical formula from changing from one mass to another

Given * (mw/mol)*ration*(mw/mol) = grams of other

Finding rations in equations
o Nitrogen and hydrogen react to form ammonia gas according to the following
equation. ___N2 + ___H2  ___NH3

If 56.0 grams of nitrogen are used up by the reaction, how many grams of
ammonia will be produced?

How many grams of hydrogen must react if the reaction needs to produce
63.5 grams of ammonia?
o Aluminum metal reacts with zinc chloride to produce zinc metal and aluminum
chloride. ___Al + ___ZnCl2  ___Zn + ___AlCl3

A mass of 45.0 grams of aluminum will react with how many grams of
zinc chloride

What mass of aluminum chloride will be produced if 22.6 grams of zinc
chloride are used up in the reaction?

Solve Stoichiometry Problems
o Given the following equation: 2 KClO3 ---> 2 KCl + 3 O2

How many moles of O2 can be produced by letting 12.00 moles of
KClO3 react?

The KClO3 / O2 molar ratio is 2/3.
o

x = 18.00 mol of O2
Given the following equation: 2 NaClO3 ---> 2 NaCl + 3 O2
12.00 moles of NaClO3 will produce how many grams of O2?


Find the limiting Reactant

Consider the following reaction: 3 Si + 2 N2 → Si3N4 a. When 21.44
moles of Si reacts with 17.62 moles of N2, how many moles of Si3N4 are
formed? b. What is the limiting reactant?

7.147 mol Si3N, Si,

21.44 mol Si * (1 mol Si3N4 / 3 mol Si) = 7.147 mol Si3N4

17.62 mol N2 * (1 mol Si3N4 / 2 mol N2) = 8.810 mol Si3N4
o Consider the following reaction: 2 CuCl2 + 4 KI → 2 CuI + 4 KCl + I2
o a. When 0.56 moles of CuCl2 reacts with 0.64 moles of KI, how many moles of
I2 are formed?

o
b. What is the limiting reactant?


0.16 mol I2
KI
Theoretical Yield/ percent yield
o If 4.50 g of HCl are reacted with 15.00 g of CaCO3, according to the following
balanced chemical equation, calculate the theoretical yield of CO2. 2HCl +
CaCO3 → CaCl 2 + H2O + CO2

1. Determine the number of moles of CO2 produced if all of each reactant
is used up

4.50 g HCl x (1mol/ 36.5g HCL)* (1mol / 2 mol) = .0616 mol
CO2


15g CaCO3 * (1 mol/ 100.1 g CACO3) * (1 /1) =.1499 mol CO2
Calculate theoretical yield of product? (Use the smallest number of moles
of the product (CO2) from step 1 to calculate the theoretical yield of
product (CO2)).


.0161 mol CO2 * (44g / 1 mol CO2) = 2.71 g CO2
If given the yield from the lab was 2.50g of CO2, what is the percent
yield?


2.50 / 2.71 * 100 = 92.3% yield
In the following reaction, 0.157g of p-acetaminophenol was used to react with 0.486 g of
acetic anhydride to produce acetaminophen and acetic acid. The product was purified
and acetimophen was extracted. The actual mass of acetaminophen produced was 0.198
g. Determine the theoretical yield and the percent yield for acetimphen.
p-Aminophenol
+
Acetic anhydride
C6H7NO

à
C4H6O3
Acetaminophen
C8H9NO2
molar mass of p-aminophenol =109.1g/mol
molar mass of acetic anhydride = 102.1 g/mol
moles of p-aminophenol = mass/molar mass
= 0.157g/(109.1g/mol)
= 0.00144 mol Limiting Reactant
moles of acetic anhydride = mass/molar mass
= 0.486g/(102.1g/mol)
= 0.00476 mol
Theoretical Yield= 0.00144 mol x 151.2g/mol
= 0.217 g

Percent yield= 0.198g
0.217g
x 100 %
= 91.2 %
+
Acetic acid
CH3COOH