Question

Balance the chemical equation given below, and calculate the volume of nitrogen monoxide gas produced when 8.00 g of ammonia is reacted with 12.0 g of oxygen at 25°C? The density of nitrogen monoxide at 25°C is 1.23 g/L. ___ NH3(g) + ___ O2(g) → ___ NO(g) + ___ H2O(l)

Answer 1

Answer:

$4NH_{3} + 5O_{2}$ → $4NO_ + 6H_{2}O$

Volume of NO = 11.46 lit

Explanation:

From the above written equation we can easily understood that $NH_{3}$ here acts as an limiting reagent.

4 mol of $NH_{3}$ can produce 4 mol of $NO$

Molecular weight of $NH_{3}$ = 17 g/mol

So, 8 g of $NH_{3}$ means = $\frac{8}{17}$ = 0.470 mol

So, 0.470 mol of $NH_{3}$ will produce 0.470 mol of NO

Molecular weight of NO = 30 g/mol

So, 0470 mol of NO means (30 × 0.470 ) = 14.1 g

we know that, Density = $\frac{Mass}{Volume}$

 Volume of NO = $\frac{14.1}{1.23}$ = 11.46 lit