Question

Copper reacts with nitric acid via the following equation:

3 Cu(s) + 8 HNO₃(aq) →

3 Cu(NO₃)₂(aq) + 2 NO(g) + 4 H₂O(ℓ)

What mass of NO(g) can be formed when 31.5 g of Cu reacts with 88.0 g HNO₃?

Answer 1

Molar mass of Cu = 63.55 g/mol

mass(Cu)= 31.5 g

number of mol of Cu,

n = mass of Cu/molar mass of Cu

=(31.5 g)/(63.55 g/mol)

= 0.4957 mol

Molar mass of HNO3,

MM = 1*MM(H) + 1*MM(N) + 3*MM(O)

= 1*1.008 + 1*14.01 + 3*16.0

= 63.018 g/mol

mass(HNO3)= 88.0 g

number of mol of HNO3,

n = mass of HNO3/molar mass of HNO3

=(88.0 g)/(63.018 g/mol)

= 1.396 mol

3 mol of Cu reacts with 8 mol of HNO3

for 0.4957 mol of Cu, 1.3218 mol of HNO3 is required

But we have 1.3964 mol of HNO3

so, Cu is limiting reagent

we will use Cu in further calculation

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

According to balanced equation

mol of NO formed = (2/3)* moles of Cu

= (2/3)*0.4957

= 0.3304 mol

mass of NO = number of mol * molar mass

= 0.3304*30.01

= 9.92 g

Answer: 9.92 g