Question

Copper reacts with nitric acid via the equation below. What mass
of NO (g) can be formed when 30.6 g of Cu reacts with 85.6 g HNO3?
Which is the limiting reagent?

Cu (s) + HNO3 (aq) ------> Cu(NO3)2 (aq) + NO (g) + H2O

What is the % yield if 7.12 g of NO was obtained experimentally?

Answer 1

Balanced equation

3 Cu (s) + 8 HNO₃ (aq) ------>3 Cu(NO₃)₂ (aq) + 2 NO (g) + 4 H₂O

molar mass of Cu = 63.55 g/mol

30.6 g / 63.55 g/mol => 0.4815 mol of Cu

Molar mass of HNO3 = 63.0 g

85.6 g / 63.0 g/mol => 1.359 mol of HNO3

From balanced equation it reaveals that 3 mol of Cu needs 8 mol of HNO₃, So 0.4815 mol of Cu will need 0.4815 mol * (8/3) => 1.284 mol of HNO3

But available amount of HNO3 is 1.359 mol ( It is excess in amount).

So Cu is limiting reagent.

3 mol of Cu will produce 2 mol of NO

Therefore 0.4815 mol of Cu will produce 0.4815*(2/3) => 0.321 mol of NO

molar mass of NO = 30.0 g/mol

Mass of NO = 0.321 mol * 30.0 g/mol => 9.63 g.

% yield if it produce 7.12 g => (7.12 g / 9.63 g ) *100 => 73.9 %