Question

Given that the relative molecular mass [molecular weight] 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 : $3CuO+2N{H}_3⟶3Cu+3H_{2}O+N_2$.

• A. $44.8L$
• B. $5.56L$
• C. $11.2L$
• D. $22.4L$

Answer 1

The correct option is D $22.4L$

$3CuO+2N{H}_3⟶3Cu+3H_{2}O+N_2$

3 mole 2 mole

3$\times$80 g 2$\times$22.4 L

= 240 g = 44.8 lit

For $240g$ of $CuO$, ammonia consumed is $44.8L$

For $120g$of $CuO$, ammonia consumed = $\frac{44.8}{2}$= $22.4L$