Question

If 3.35 g Cu(NO3)2 are obtained from allowing 1.46g of Cu to react with excess HNO3--

How many grams of copper nitrate are formed from 1.46 g copper?

What is the percent yield of copper nitrate for this reaction?

Answer 1

1)

Molar mass of Cu = 63.55 g/mol

mass(Cu)= 1.46 g

use:

number of mol of Cu,

n = mass of Cu/molar mass of Cu

=(1.46 g)/(63.55 g/mol)

= 2.297*10^-2 mol

Balanced chemical equation is:

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

Molar mass of Cu(NO3)2,

MM = 1*MM(Cu) + 2*MM(N) + 6*MM(O)

= 1*63.55 + 2*14.01 + 6*16.0

= 187.57 g/mol

According to balanced equation

mol of Cu(NO3)2 formed = (1/1)* moles of Cu

= (1/1)*0.023

= 0.023 mol

use:

mass of Cu(NO3)2 = number of mol * molar mass

= 2.297*10^-2*1.876*10^2

= 4.309 g

Answer: 4.31 g

2)

% yield = actual mass*100/theoretical mass

= 3.35*100/4.3092

= 77.7 %

Answer: 77.7 %