Question

How many grams of CuO would be required to produce 10.0 moles of nitrogen gas in the following chemical reaction?
2 NH₃ (g) + 3 CuO (s) → 3 Cu (s) + N₂ (g) + 3 H₂O (g)

Answer 1

According to equation mole ratio $n(Cu):n(N_2)=3:1$ , then $n(Cu) = 3\times n(N_2) = 3\times 10.0 = 30.0$ mol

$m=M\times n$

$M(CuO) =79.55\frac{g}{mol}$

$m(CuO) = 30.0\times 79.55 = 2386.5 g = 2390 g$