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

Question

Quizplus · Accepted Answer

The balanced chemical equation is:3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂OFrom the stoichiometry of the balanced chemical equation, we can see 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 of 0.602 M HNO₃) x (1 L / 1000 mL) x (0.602 mol HNO₃ / L) x (3 mol Cu / 8 mol HNO₃) = 0.00279 mol CuThe mass of Cu in the sample is:mass of Cu = (0.00279 mol Cu) x (63.55 g Cu / mol Cu) = 0.177 g CuTherefore, the percent copper (by mass) in the sample is:percent copper = (0.177 g Cu / 0.411 g sample) x 100% = 43.1%

A 0411 G Sample of Powdered Copper Mixed with an Inert,soluble