Question

Cu(s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)

Answer 1

Answer:

The percent by mass of copper in the mixture was 32%

Explanation:

The ammount of HNO₃ used is:

mol HNO₃ = volume * concentration

mol HNO₃ = 0.015 l * 15.8 mol/l

mol HNO₃ = 0.237 mol

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%

Answer 2

Answer:

B) 32%

Explanation:

The data from the problem of which we are interested:

2 gram sample

0.010 moles of Cu(NO₃)₂ produced by the reaction

we calculate the mass of Cu(NO₃)₂ knowing the molecular weight of copper nitrate which is equal to 187.5 grams/mole:

mass = number of moles × molecular weight

mass =  0.010 × 187.5  = 1.875 g

now we devise the following reasoning:

if         187.5 grams of copper nitrate contains 63.5 grams of cooper

then    1.875 grams of copper nitrate contains X grams of cooper

X = (1.875 ×  63.5) /  187.5 = 0.645 grams of cooper

And we have all the values to calculate the percent of Cu by mass in the original mixture:

if in            2 grams of sample we have 0.645 grams of cooper

then in  100  grams of sample we have Y grams of cooper

Y = (100 ×  0.645) / 2 = 31.75 ≈ 32 %