Question

Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120g of copper oxide? The equation for the reaction is:

\[\ce{3CuO + 2NH3 → 3Cu + 3H2O + N2}\]

Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at S.T.P.) is required to completely reduce 120 g of copper oxide? The equation for the reaction is:

\[\ce{3CuO + 2NH3 → 3Cu + 3H2O + N2}\]

(Volume occupied by 1 mole of gas at S.T.Р. is 22.4 litres).

Answer 1

Balanced Equation:

\[\ce{3CuO + 2NH3 → 3Cu + 3H2O + N2}\]

Mass of 3 moles of CuO = 3 × 80 = 240 g

Volume of 2 moles of NH₃ at STP = 2 × 22.4 = 44.8 L (or dm³)

Calculation:

∵ 240 g of CuO requires = 44.8 L of NH₃

∴ 120 g of CuO will require = `44.8/240 xx 120`

= 22.4 L

The volume of ammonia required is 22.4 L.