Question

2CuO+2NH3------ 3Cu + 3H2O+ N2. Given that the relative molecular mass of copper oxide is 80, what volume of ammonia is required to completely reduce 120 g of Copper oxide? ( Cu=64, O=16, N=14)​

Answer 1

Volume of Ammonia(NH₃) = 22.4 L

Further explanation

Given

Reaction

3CuO+2NH₃⇒ 3Cu + 3H₂O+ N₂

In the problem, the CuO coefficient should be 3 not 2

M CuO = 80

mass CuO = 120 g

Required

The volume of NH₃

Solution

mol CuO :

`\tt mol=(mass)/(M)\\\\mol=(120)/(80)\\\\mol=1.5`

From the equation, mol ratio CuO : NH₃ = 3 : 2, so mol NH₃=

`\tt (2)/(3)* 1.5=1~mol`

Assume at STP(0 °C, 1 atm) ⇒1 mol = 22.4 L, then volume of NH₃=22.4 L