A 0.411 g sample of powdered copper mixed with an inert, soluble additive was fully consumed by 23.4 mL of 0.602 M nitric acid, producing copper(II) nitrate, water, and nitric oxide (NO) gas. What was the percent copper (by mass) in the sample? 3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O A) 81.7% B) 68.3% C) 65.3% D) 56.9% E) 34.4%

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 3 moles of Cu react with 8 moles of HNO3. Therefore, the number of moles of Cu in the sample is:```moles of Cu = (23.4 mL * 0.602 M) * (3 moles Cu / 8 moles HNO3) = 0.00539 moles Cu```The mass of Cu in the sample is:```mass of Cu = 0.00539 moles Cu * 63.55 g/mol = 0.343 g Cu```The percent copper (by mass) in the sample is:```% Cu = (0.343 g Cu / 0.411 g sample) * 100% = 83.5%```Therefore, the correct answer is A) 81.7%.

A 0411 G Sample of Powdered Copper Mixed with an Inert