When [tex]8.00 \times 10^{22}[/tex] molecules of ammonia react with [tex]7.00 \times 10^{22}[/tex] molecules of oxygen according to the chemical equation shown below, how many grams of nitrogen gas are produced?
\[ 4\text{NH}_3(g) + 3\text{O}_2(g) \rightarrow 2\text{N}_2(g) + 6\text{H}_2\text{O}(g) \]
Answer (2)
NH₃:
N = 8 10²² NA = 6.02 10²³
n = N/NA = 8 10²²/6.02 10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7 10²² NA = 6.02 10²³
n = N/NA = 7 10²²/6.02 10²³ = 1.16*10⁻¹=0.116mol
4NH₃ + 3O ₂ ⇒ 2N ₂ + 6H ₂ O
4mol : 3mol : 2mol 0.133mol : 0.116mol : 0,0665mol limiting reactant
N₂:
n = 0.0665mol M = 28g/mol
m = n M = 0.0665mol 28g/mol = 1,862g
When 8.00 × 1 0 22 molecules of ammonia react with 7.00 × 1 0 22 molecules of oxygen, about 1.86 grams of nitrogen gas (N₂) are produced. The limiting reactant is ammonia (NH₃). The conversion relies on the stoichiometry of the balanced chemical equation.
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