Question

1.If you started with 3.60g of Cu and reacted it with an excess of nitric acid and water, what mass of copper hydroxide would you expect to recover?

2.If you recovered only 64.3% of the expected copper hydroxide, how much would you have recovered?

3.If you started a experiment with 2.9g of Cu and every other reactant was in excess, how much Zn would you need to complete this experiment?

Answer 1

Q1.

m = 3.6 g of Cu

Cu + 2HNO3 --> Cu+2(aq) + H2(g) + 2NO3-(aq)

mol of Cuy = mass/MW = 3.6/63.5 = 0.056692 mol of Cu(s)

so... 1 mol of Cu = 1 mol of Cu+2

0.056692 mol of Cu+2 forms

1 mol of Cu+2 + 2(OH-) = Cu(OH)2

so

0.056692 mol of Cu(OH)2

MW of Cu(OH)2 = 97.561

mass of Cu(OH)2 = 97.561 *0.056692 = 5.530 g of Cu(OH)2 could be obtained

Q2.

64.3% only.. then --> 5.530*64.3/100 = 3.55579 g of Cu(OH)2

Q3.

m = 2.9 g of Cu ...

Zn for this experiment

ratio is 1:1

mol o Cu = mass/MW = 2.9/63.5 = 0.0456 mol of Cu

so

0.0456 mol of Zn required

mass of Zn = mol*MW = 65.38 *0.0456 = 2.9813 g of Zn