What is the percent yield of copper when 14.0g of Cu2O reacts with Cu2S and 12.3 grams of copper is recovered? Round to two decimal places and do not use a percent sign in your answer.
2Cu2O + Cu2S → 6Cu + SO2
What is 12.3/14.0 * 100. Round to two decimal places. 85.86?
To find the percent yield of copper in this reaction, we first need to calculate the theoretical yield of copper. The theoretical yield is the maximum amount of copper that can be obtained from the given amount of reactant.
1. Calculate the molar mass of Cu2O:
Cu: 2 atoms x atomic mass of Cu = 2 x 63.55 g/mol = 127.1 g/mol
O: 1 atom x atomic mass of O = 1 x 16.00 g/mol = 16.00 g/mol
Total molar mass = 127.1 g/mol + 16.00 g/mol = 143.1 g/mol
2. Convert the given mass of Cu2O to moles:
Moles of Cu2O = 14.0 g / 143.1 g/mol ≈ 0.0977 mol
3. Use the balanced equation to determine the mole ratio between Cu2O and Cu:
From the balanced equation: 2 moles of Cu2O produce 6 moles of Cu
Therefore, 0.0977 mol of Cu2O produces (6/2) x 0.0977 mol of Cu = 0.2931 mol of Cu
4. Calculate the molar mass of Cu:
Cu: 1 atom x atomic mass of Cu = 1 x 63.55 g/mol = 63.55 g/mol
5. Convert the calculated moles of Cu to grams:
Grams of Cu = 0.2931 mol x 63.55 g/mol ≈ 18.63 g
Now, we can calculate the percent yield of copper.
6. Calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) x 100
Actual yield = 12.3 g
Theoretical yield = 18.63 g
Percent yield = (12.3 g / 18.63 g) x 100 ≈ 66.07
Therefore, the percent yield of copper in this reaction, rounded to two decimal places, is approximately 66.07. Note that no percent sign is included in the answer.