25.00 mL of 5.2159 M sulfuric acid (aq) is mixed with 25.00 mL 3.9009 M barium hydroxide (aq).
1. Write the balanced chemical equation. What is the precipitate?
H2SO4 (aq) + Ba(OH)2 (aq) à BaSO4 (s) + 2 H2O (l)
Barium sulfate is the precipitate.
2. What is the limiting reactant?
Determine number of moles of reactants:
5.2159 M H2SO4 * 0.02500 L = 0.1304 mol H2SO4
3.9009 M Ba(OH)2 * 0.02500 L = 0.09752 mol Ba(OH)2
There is more H2SO4 than Ba(OH)2. Since they react in a 1:1 ratio, Ba(OH)2 is therefore the limiting reactant.
3. How much precipitate is formed (in g)? What is the mass of excess reagent?
All of the Ba(OH)2 will be used up. Since molar ratio for Ba(OH)2 and BaSO4 is 1:1, the moles of BaSO4 product will be 0.09752 mol. Multiply by the molar mass of barium sulfate.
0.09752 mol BaSO4 * 233.3984 g/1 mol = 22.76 g BaSO4 (s)
0.09752 mol of H2SO4 will react with the same number of moles of barium hydroxide. Originally, there was 0.1304 mol H2SO4. Multiply the remaining number of moles by the molar mass of sulfuric acid.
0.1304 mol H2SO4 – 0.09752 mol H2SO4 = 0.03288 mol H2SO4
0.03288 mol H2SO4 * 98.0842 g/1 mol = 3.225 g H2SO4 (aq)
4. If you weigh 18.24 g precipitate, what was the percent yield?
% yield = actual yield / theoretical yield = 18.24 g/22.76 g = 80.14%
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