When excess dry ammonia gas is passed over 24.0\text{ g} of heated copper(II) oxide (CuO),… — JAMB UTME

When excess dry ammonia gas is passed over

24.0\text{ g}

of heated copper(II) oxide (

CuO

), the oxide is completely reduced to copper metal according to the equation:

2NH_3(g) + 3CuO(s) \rightarrow 3Cu(s) + N_2(g) + 3H_2O(l)

What is the volume of nitrogen gas, in

\text{dm}^3

, evolved at s.t.p.?
(

Cu = 64.0

O = 16.0

, molar volume of gas at s.t.p.

= 22.4\text{ dm}^3\text{mol}^{-1}

)

Answer: 2.24 dm³

The volume of nitrogen gas evolved at s.t.p. is

2.24\text{ dm}^3

According to the balanced chemical equation

2NH_3(g) + 3CuO(s) \rightarrow 3Cu(s) + N_2(g) + 3H_2O(l)

, 3 moles (

240.0\text{ g}

) of copper(II) oxide produce 1 mole (

22.4\text{ dm}^3

at s.t.p.) of nitrogen gas. Hence,

24.0\text{ g}

CuO

yields

\frac{24.0}{240.0} \times 22.4\text{ dm}^3 = 2.24\text{ dm}^3

of nitrogen gas.

Calculate the molar mass of

CuO

Molar mass of

CuO = 64.0 + 16.0 = 80.0\text{ g mol}^{-1}

Required to convert the given mass of reactant to moles.

Find the number of moles of

CuO

Moles of

CuO = \frac{24.0\text{ g}}{80.0\text{ g mol}^{-1}} = 0.30\text{ mol}

Determines the exact amount of copper(II) oxide reacting.

Apply stoichiometry to find moles of

N_2

gas produced

Moles of

N_2 = 0.30\text{ mol } CuO \times \frac{1\text{ mol } N_2}{3\text{ mol } CuO} = 0.10\text{ mol } N_2

The mole ratio between

CuO

and

N_2

in the balanced chemical equation is

3:1

Convert moles of

N_2

to volume at s.t.p.

Volume of

N_2 = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

at s.t.p.

Reduction of metallic oxides by ammonia gas and gas stoichiometry at STP

Question