Question

Given the following equation: 4 NH3 (g)5 O2 (g) >4 NO (g) + 6 H20 (I) How many moles of NH3 is required to react with 25.7 grams of O2?

Answer 1

Answer:

0.6425 moles of $NH_3$ is required to react with 25.7 grams of $O_2$.

Explanation:

mas of oxygen gas = 25.7 g

moles of oxygengas = $\frac{25.7 g}{32 g/mol}=0.8031 mol$

$4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)$

According to reaction given above,  5 moles of oxygen gas reacts with 4 moles of ammonia gas.

Then 0.8031 moles of oxygen gas will react with :

$\frac{4}{5}\times 0.8031 mol=0.6425 mol$ of ammonia gas

0.6425 moles of $NH_3$ is required to react with 25.7 grams of $O_2$.