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 :

$\tt mol=\dfrac{mass}{M}\\\\mol=\frac{120}{80}\\\\mol=1.5$

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

$\tt \dfrac{2}{3}\times 1.5=1~mol$

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