Chemistry: Stoichiometry
Mole ratios, limiting reactants, yields, and solution stoichiometry
Chemistry: Stoichiometry
Mole ratios, limiting reactants, yields, and solution stoichiometry
Chemistry - Grade advanced
- 1
Balance the equation and determine the mass of oxygen gas needed to completely react with 12.0 g of propane: C3H8 + O2 -> CO2 + H2O.
Use the mole ratio 1 mol C3H8 to 5 mol O2 after balancing the equation.
The balanced equation is C3H8 + 5O2 -> 3CO2 + 4H2O. 12.0 g C3H8 is 0.272 mol C3H8, so 1.36 mol O2 is needed. The required mass of oxygen gas is 43.5 g O2. - 2
Aluminum reacts with chlorine gas to form aluminum chloride: 2Al + 3Cl2 -> 2AlCl3. If 5.40 g Al reacts with 8.10 g Cl2, identify the limiting reactant and calculate the theoretical mass of AlCl3 produced.
Convert both reactants to moles, then compare the available mole ratio to the balanced equation ratio.
5.40 g Al is 0.200 mol Al, and 8.10 g Cl2 is 0.114 mol Cl2. The chlorine gas is limiting because it can react with only 0.0762 mol Al. The reaction produces 0.0762 mol AlCl3, which is 10.2 g AlCl3. - 3
Iron(III) oxide reacts with carbon monoxide in a blast furnace: Fe2O3 + 3CO -> 2Fe + 3CO2. How many grams of iron can be produced from 250.0 g Fe2O3 if carbon monoxide is in excess?
250.0 g Fe2O3 is 1.565 mol Fe2O3. The mole ratio gives 3.130 mol Fe. Multiplying by 55.85 g/mol gives 174.8 g Fe, so 174.8 g of iron can be produced. - 4
A reaction has a theoretical yield of 28.6 g but produces 21.4 g of product in the lab. Calculate the percent yield.
Percent yield compares what was actually made to what could have been made under ideal conditions.
Percent yield equals actual yield divided by theoretical yield times 100. The percent yield is 21.4 g divided by 28.6 g times 100, which equals 74.8%. - 5
Magnesium nitride reacts with water: Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3. How many grams of ammonia can form from 18.5 g Mg3N2 and 14.0 g H2O?
First find the limiting reactant, then use that reactant to calculate the ammonia produced.
18.5 g Mg3N2 is 0.184 mol Mg3N2, and 14.0 g H2O is 0.777 mol H2O. The water is limiting because 0.184 mol Mg3N2 would require 1.10 mol H2O. The reaction produces 0.259 mol NH3, which is 4.41 g NH3. - 6
For the reaction N2 + 3H2 -> 2NH3, a mixture contains 4.00 mol N2 and 9.00 mol H2. Determine the limiting reactant, the moles of NH3 produced, and the moles of excess reactant remaining.
Hydrogen is the limiting reactant because 9.00 mol H2 can react with only 3.00 mol N2. The reaction produces 6.00 mol NH3. Since 3.00 mol N2 reacts, 1.00 mol N2 remains in excess. - 7
Calcium carbonate decomposes when heated: CaCO3 -> CaO + CO2. If 75.0 g CaCO3 is heated and 28.0 g CO2 is collected, what is the percent yield of CO2?
The mole ratio between CaCO3 and CO2 is 1 to 1.
75.0 g CaCO3 is 0.749 mol CaCO3, so the theoretical amount of CO2 is 0.749 mol. This equals 33.0 g CO2. The percent yield is 28.0 g divided by 33.0 g times 100, which equals 84.8%. - 8
Silver nitrate reacts with sodium chloride: AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq). What volume of 0.250 M AgNO3 is required to completely react with 35.0 mL of 0.180 M NaCl?
Use molarity times volume in liters to find moles of NaCl.
35.0 mL of 0.180 M NaCl contains 0.00630 mol NaCl. The reaction ratio is 1 to 1, so 0.00630 mol AgNO3 is required. The volume of 0.250 M AgNO3 needed is 0.0252 L, or 25.2 mL. - 9
Hydrochloric acid neutralizes barium hydroxide: 2HCl + Ba(OH)2 -> BaCl2 + 2H2O. How many milliliters of 0.150 M HCl are needed to neutralize 20.0 mL of 0.100 M Ba(OH)2?
20.0 mL of 0.100 M Ba(OH)2 contains 0.00200 mol Ba(OH)2. The balanced equation requires 2 mol HCl for every 1 mol Ba(OH)2, so 0.00400 mol HCl is needed. The required volume is 0.0267 L, or 26.7 mL HCl. - 10
Methane combusts according to CH4 + 2O2 -> CO2 + 2H2O. If 6.50 L CH4 reacts with excess oxygen at the same temperature and pressure, what volume of CO2 forms?
For gases under the same conditions, coefficients can be used as volume ratios.
At the same temperature and pressure, gas volume ratios match mole ratios. The ratio of CH4 to CO2 is 1 to 1, so 6.50 L of CO2 forms. - 11
Ammonia burns in oxygen: 4NH3 + 5O2 -> 4NO + 6H2O. If 12.0 L NH3 and 20.0 L O2 react at the same temperature and pressure, which gas is limiting and what volume of NO forms?
Use gas volume ratios from the balanced equation because all gases are measured under the same conditions.
12.0 L NH3 requires 15.0 L O2, so oxygen is in excess and ammonia is limiting. The mole ratio of NH3 to NO is 1 to 1, so 12.0 L of NO forms. - 12
Potassium chlorate decomposes: 2KClO3 -> 2KCl + 3O2. What mass of KClO3 is required to produce 15.0 L O2 at STP?
At STP, 15.0 L O2 is 0.670 mol O2. The mole ratio requires 2 mol KClO3 for every 3 mol O2, so 0.446 mol KClO3 is needed. Multiplying by 122.55 g/mol gives 54.7 g KClO3. - 13
A 2.50 g sample of impure zinc reacts with excess hydrochloric acid: Zn + 2HCl -> ZnCl2 + H2. If 0.765 L H2 is collected at STP, what is the percent zinc by mass in the sample?
Find moles of hydrogen gas first, then use the 1 to 1 mole ratio to find the mass of zinc that reacted.
0.765 L H2 at STP is 0.0342 mol H2. The mole ratio of Zn to H2 is 1 to 1, so the sample contained 0.0342 mol Zn. This is 2.23 g Zn, so the percent zinc by mass is 2.23 g divided by 2.50 g times 100, or 89.2%. - 14
Phosphoric acid reacts with sodium hydroxide: H3PO4 + 3NaOH -> Na3PO4 + 3H2O. How many grams of Na3PO4 can form when 50.0 mL of 0.200 M H3PO4 reacts with 75.0 mL of 0.500 M NaOH?
The solution contains 0.0100 mol H3PO4 and 0.0375 mol NaOH. To use all the acid, 0.0300 mol NaOH is required, so H3PO4 is limiting. The reaction forms 0.0100 mol Na3PO4, which is 1.64 g Na3PO4. - 15
Octane combusts according to 2C8H18 + 25O2 -> 16CO2 + 18H2O. If 10.0 g C8H18 burns in 40.0 g O2, calculate the theoretical mass of CO2 produced and identify the excess reactant.
Use the octane to oxygen ratio 2 to 25, then use the limiting reactant to calculate carbon dioxide.
10.0 g C8H18 is 0.0875 mol octane, and 40.0 g O2 is 1.25 mol O2. Complete combustion of the octane would require 1.09 mol O2, so oxygen is in excess and octane is limiting. The reaction produces 0.700 mol CO2, which is 30.8 g CO2.