Moles and stoichiometry practice problems (from Chapter 3 in Brady, Russell, and Holum’s Chemistry, Matter and its Changes, 3rd Ed.)
1) How many moles of sodium atoms correspond to 1.56x1021 atoms of sodium?
2) How many moles of Al atoms are needed to combine with 1.58 mol of O atoms to make aluminum oxide, Al2O3?
3) How many moles of Al are in 2.16 mol of Al2O3?
4) Aluminum sulfate, Al2(SO4)3, is a compound used in sewage treatment plants.
a. Construct a pair of conversion factors that relate moles of aluminum to moles of sulfur for this compound
b. Construct a pair of conversion factors that relate moles of sulfur to moles of Al2(SO4)3
c. How many moles of Al are in a sample of this compound if the sample also contains 0.900 mol S?
d. How many moles of S are in 1.16 mol Al2(SO4)3?
5) How many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3?
6) What is the total number of atoms in 0.260 mol of glucose, C6H12O6?
7) What is the mass of 1.00 mol of each of the following elements?
8) Determine the mass in grams of each of the following:
a. 1.35 mol Fe
b. 24.5 mol O
c. 0.876 mol Ca
d. 1.25 mol Ca3(PO4)2
e. 0.625 mol Fe(NO3)3
f. 0.600 mol C4H10
g. 1.45 mol (NH4)2CO3
9) Calculate the number of moles of each compound:
a. 21.5 g CaCO3
b. 1.56 g NH3
c. 16.8 g Sr(NO3)2
d. 6.98 mg Na2CrO4
10) Calculate the percentage composition by mass of each element in the following compounds:
11) Phencyclidine is C17H25N. A sample suspected of being this illicit drug was found to have a percentage composition of 83.71% C, 10.42% H, and 5.61% N. Do these data acceptably match the theoretical data for phencyclidine?
12) How many grams of O are combined with 7.14x1021 atoms of N in the compound N2O5?
13) Quantitative analysis of a sample of sodium pertechnetate with a mass of 0.896g found 0.111g Na and 0.477g technetium (Tc). The remainder was oxygen. Calculate the empirical formula of sodium pertechnetate, NaxTcyOz.
14) A substance was found to be composed of 22.9% Na, 21.5% B, and 55.7% O. What is the empirical formula of this compound?
15) When 0.684 g of an organic compound containing only C, H, and O was burned in oxygen 1.312g CO2 and 0.805g H2O were obtained. What is the empirical formula of the compound?
16) Write the equation that expresses in acceptable chemical shorthand the following statement: “Iron can be made to react with molecular oxygen (O2) to give iron oxide with the formula Fe2O3”
17) Balance the following reactions:
a. Ca(OH)2 + HCl à CaCl2 + H2O
b. AgNO3 + CaCl2 à Ca(NO3)2 +AgCl
c. Fe2O3 + C à Fe + CO3
d. NaHCO3 + H2SO4 à Na2SO4 + H2O + CO2
e. C4H10 + O2 à CO2 +H2O
f. Mg(OH)2 + HBr à MgBr2 + H2O
g. Al2O3 + H2SO4 à Al2(SO4)3 + H2O
h. KHCO3 + H3PO4 à K2HPO4 + H2O + CO2
i. C9H10O + O2 à CO2 + H2O
18) Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate, Na2S2O3:
Na2S2O3(aq) + 4Cl2(g) + 5H2O(aq) à 2NaHSO4(aq) + 8HCl(aq)
a. How many moles of Na2S2O3 are needed to react with 0.12mol of Cl2?
b. How many moles of HCl can form from 0.12mol of Cl2?
c. How many moles of H2O are required for the reaction of 0.12mol of Cl2?
d. How many moles of H2O react if 0.24mol HCl is formed?
19) The incandescent white of a fireworks display is caused by the reaction of phosphorous with O2 to give P4O10.
a. Write the balanced chemical equation for the reaction.
b. How many grams of O2 are needed to combine with 6.85g of P?
c. How many grams of P4O10 can be made from 8.00g of O2?
d. How many grams of P are needed to make 7.46g P4O10?
20) In dilute nitric acid, HNO3, copper metal dissolves according to the following equation:
3Cu(s) + 8HNO3(aq) à 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(aq)
How many grams of HNO3 are needed to dissolve 11.45g of Cu?
21) The reaction of powdered aluminum and iron(II)oxide,
2Al(s) + Fe2O3(s) à Al2O3(s) + 2Fe(l)
produces so much heat the iron that forms is molten. Because of this, railroads use the reaction to provide molten steel to weld steel rails together when laying track. Suppose that in one batch of reactants 4.20mol Al was mixed with 1.75mol Fe2O3.
a. Which reactant, if either, was the limiting reactant?
b. Calculate the mass of iron (in grams) that can be formed from this mixture of reactants.
22) Silver nitrate, AgNO3, reacts with iron(III) chloride, FeCl3, to give silver chloride, AgCl, and iron(III) nitrate, Fe(NO3)3. A solution containing 18.0g AgNO3 was mixed with a solution containing 32.4g FeCl3. How many grams of which reactant remains after the reaction is over?
23) Barium sulfate, BaSO4, is made by the following reaction:
Ba(NO3)2(aq) + Na2SO4(aq) à BaSO4(s) + 2NaNO3(aq)
An experiment was begun with 75.00g of Ba(NO3)2 and an excess of Na2SO4. After collecting and drying the product, 63.45g BaSO4 was obtained. Calculate the theoretical yield and percent yield of BaSO4.
24) Aluminum sulfate can be made by the following reaction:
2AlCl3(aq) + 3H2SO4(aq) à Al2(SO4)3(aq) + 6HCl(aq)
It is quite soluble in water, so to isolate it the solution has to be evaporated to dryness. This drives off the volatile HCl, but the residual solid has to be treated to a little over 200°C to drive off all the water. In one experiment, 25.0g of AlCl3 was mixed with 30.0g H2SO4. Eventually, 28.46g of pure Al2(SO4)3 was isolated. Calculate the percent yield.
1) 2.59x10-3mol Na atoms
2) 1.05mol Al
3) 4.32mol Al
4) a. 2mol Al/3mol S b. 3mol S/1mol Al2(SO4)3 c. 0.600mol Al d. 3.48mol S
5) 0.0725mol N2 and 0.218mol H2
6) 3.76x1024 atoms
7) a. 23.0g Na b. 32.1g S c. 35.3g Cl
8) a. 75.4g Fe b. 392g O c. 35.1g Ca d. 388g Ca3(PO4)2
e. 151g Fe(NO3)2 f. 34.9g C4H10 g. 139g (NH4)2CO3
9) a. 0.215mol b. 0.0916mol c. 0.0794mol d. 4.31x10-8mol
10) a. 19.2% Na, 1.68% H, 25.8% P, 53.3% O
b. 12.2% N, 5.26% H, 26.9% P, 55.6%O
c. 62.0% C, 10.4% H, 27.6% O
11) Theoretical data (83.89% C, 10.35% H, 5.76% N) are consistent with experimental results.
12) 0.474g O
16) 4Fe + 3O2 à 2Fe2O3
a. Ca(OH)2 + 2HCl à CaCl2 + 2H2O
b. 2AgNO3 + CaCl2 à Ca(NO3)2 + 2AgCl
c. 2Fe2O3 + 3C à 4Fe + 3CO3
d. 2NaHCO3 + H2SO4 à Na2SO4 + 2H2O + 2CO2
e. 2C4H10 + 13O2 à 8CO2 + 10H2O
f. Mg(OH)2 + 2HBr à MgBr2 + 2H2O
g. Al2O3 + 3H2SO4 à Al2(SO4)3 + 3H2O
h. 2KHCO3 + H3PO4 à K2HPO4 + 2H2O + 2CO2
i. C9H10O + 14O2 à 9CO2 + 10H2O
18) a. 0.030mol Na2S2O3 b. 0.24mol HCl c. 0.15mol H2O
d. 0.15mol H2O
19) a. 4P + 5O2 à P4O10 b. 8.85g O2 c. 14.2g P4O10 d. 3.26g P
20) 30.31g HNO3
21) a. limiting reactant is Fe2O3 b. 195g Fe is formed
22) 26.7g of FeCl3 are left over
23) theoretical yield = 66.98g BaSO4, % yield = 94.73%
24) % yield = 88.74%