Mole Calculations Study Guide 2
Name: Key
Gram to Gram Conversions
1. How many grams of Zinc are needed to completely react with 3.2g sulfur, producing
zinc sulfide? (Hint: start by writing the balanced equation for the reaction)
Zn + S  ZnS
3.2 g S x 1 mol S/32 g = 0.10 mol S
0.10 mol S x 1 mol Zn/1 mol S = 0.10 mol Zn
0.10 mol Zn x 65.37 g Zn/1 mol Zn = 6.5 g Zn
2. Consider the following reaction,
2Cu
+
S
Cu2S
If you have 6.35 grams of copper, how many grams of sulfur would you need to
react completely?
6.35 g Cu x 1 mol Cu/63.55 g Cu = 0.10 mol Cu
0.10 mol Cu x 1 mol S/2 mol Cu = 0.050 mol S
0.050 mol S x 32.07 g S/1 mol S = 1.6 g S
3. How many grams of nitrogen are needed to react with 3.20 grams of oxygen to
create nitrogen dioxide?
N2 + 2O2  2NO2
3.20
g O2 x 1 mol O2/32.00 g O2 = 0.10 mol O2
0.10 mol O2 x 1 mol N2/2 mol O2 = 0.050 mol N2
0.050 mol N2 x 28.02 g N2/1 mol N2 = 1.40 g N2
4. How many grams of Fe2O3 are needed to react with 5.2 grams of Aluminum?
2 Al + Fe2O3  Al2O3 + 2 Fe
5.2 g Al x 1 mol Al/26.98 g Al = 0.19 mol Al
0.19 mol Al x 1 mol Fe2O3/2 mol Al = 0.095 mol Fe2O3
0.095 mol Fe2O3 x 159.70 g/mol Fe2O3 = 15.2 grams Fe2O3
5. How many grams of Silver will you get when 6.3 grams of copper reacts with excess
silver nitrate?
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
6.3 g Cu x 1 mol Cu/63.55 g Cu = 0.099 mol Cu
0.099 mol Cu x 2 mol Ag/1 mol Cu = 0.198 mol Ag
0.198 mol Ag x 107.87 g Ag/1mol Ag = 21.4 g Ag
Stoichiometry with Concentration
1. How many mL of 4.0M HCl are needed to completely react with 250mL of
6.0M NaOH?
HCl + NaOH  H2O + NaCl
0.250 L x 6.0 mol NaOH/1 L = 1.5 mol NaOH
1.5 mol NaOH x 1 mol HCl/1 mol NaOH = 1.5 mol HCl
1.5 mol HCl x 1 L/4.0 mol HCl = 0.375 L HCl or 375 mL
2. How many mL of 1.0M AgNO3 are needed to react with 2.5 g of copper?
Cu + 2AgNO3  Cu(NO3)2 + 2Ag
2.5 g Cu x 1 mol Cu/63.55 g Cu = 0.039 mol Cu
0.039 mol Cu x 2 AgNO3/1 Cu = 0.078 mol AgNO3
0.078 mol AgNO3 x 1 L/1 mol AgNO3 = 0.078 L AgNO3 or 78 mL
3. How many mL of 0.10M acetic acid are needed to react with 5.0 g of baking
soda (NaHCO3)?
NaHCO3 + HC2H3O2  NaC2H3O2 + CO2 + H2O
5.0 g NaHCO3 x 1 mol NaHCO3/84.01 g NaHCO3 = 0.0595 mol NaHCO3
0.0595 mol NaHCO3 x 1 HC2H3O2/1 mol NaHCO3 = 0.0595 mol HC2H3O2
0.0595 mol HC2H3O2 x 1 L/0.10 mol HC2H3O2 = 0.595 L or 595 mL
4. If you threw 5.0 g of potassium into 5.0 L of water, assuming that no water
was lost in the reaction, what would the concentration of the potassium
hydroxide be?
2K + 2H2O  2KOH + H2
5.0 g K x 1 mol K/39.10 g K = 0.128 mol K
0.128 mol K x 2 mol KOH/2 mol K = 0.128 mol KOH
0.128 mol KOH/5.0 L = 0.026 mol/L
5. How many mL of 2.0M CuSO4 would you need to react with 1.5 g of iron to
produce copper and iron (II) sulfate?
Fe + CuSO4  Cu + FeSO4
1.5 g Fe x 1 mol Fe/55.85 g Fe = 0.0269 mol Fe
0.0269 mol Fe x 1 mol CuSO4/1 mol Fe = 0.0269 mol CuSO4
0.0269 mol CuSO4 x 1 L/2.0 mol CuSO4 = 0.0134 L or 13.4 mL