Question

When 8.00 × 1022 molecules of ammonia react with 7.00 × 1022 molecules of oxygen according to the chemical equation shown below, how many grams of nitrogen gas are produced?
4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g)

Answer 1

Answer: 1.848 g

Explanation: To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's number}}$    ....(1)

For ammonia:

Putting values in above equation, we get:

$\text{Moles of ammonia}=\frac{8.00\times 10^{22}}{6.023\times 10^{23}}=0.132mol$

$\text{Moles of oxygen}=\frac{7.00times 10^{22}}{6.023\times 10^{23}}=0.116mol$

For the reaction:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

By Stoichiometry of the reaction,

4 moles of ammonia combine with 3 moles of Oxygen

Thus 0.132 moles of ammonia will combine with=$\frac{3}{4}\times 0.132=0.009mol$ of oxygen

Thus ammonia is the limiting reagent as it limits the formation of product.

4 moles of ammonia produces 2 moles of nitrogen

0.132 moles of ammonia will produce=$\frac{2}{4}\times 0.132=0.066 moles$ of nitrogen

Molar mass of nitrogen =  28 g/mol

Amount of nitrogen produced=${text {no of moles}}\times {text {molar mass}}=0.066\times 28=1.848g$