Cu(s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)
Each student in a class placed a
2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The students slowly poured 15.0 mL of 15.8 M HNO3 (aq) into their beakers. The reaction between copper in the mixture and HNO3 depicted above. the students observed that a brown gas was released from the beakers and that solutions turned blue, indicating the formation of Cu2+. the solutions were then diluted with distilled water to known volumes. The students determined that the reaction produced .010 mole fo Cu(NO3)2. Based on the measurement, what was the percent of Cu by mass in the original 2.00 gram sample of the mixture? (please provide detailed explanation)
A)16%
B) 32%
C) 64%
D)96%

According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.

Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.

1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:

0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.

Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:

100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g

Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%