n₂(g) + 3h₂(g) → 2nh₃(g)\nhow many moles of n₂ are needed to produce 415 l of nh₃ at stp?\n? mol n₂
n₂(g) + 3h₂(g) → 2nh₃(g)\nhow many moles of n₂ are needed to produce 415 l of nh₃ at stp?\n? mol n₂
Answer
Explanation:
Step1: Determine moles of NH₃ at STP
At STP, 1 mole of any gas occupies 22.4 L. So, $n_{NH_3}=\frac{V_{NH_3}}{V_m}$, where $V_{NH_3} = 415$ L and $V_m=22.4$ L/mol. Thus, $n_{NH_3}=\frac{415}{22.4}\approx18.53$ mol.
Step2: Use mole - ratio from balanced equation
The balanced equation is $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$. The mole - ratio of $N_2$ to $NH_3$ is $\frac{n_{N_2}}{n_{NH_3}}=\frac{1}{2}$. So, $n_{N_2}=\frac{1}{2}n_{NH_3}$. Substituting $n_{NH_3}\approx18.53$ mol, we get $n_{N_2}=\frac{1}{2}\times18.53 = 9.265$ mol.
Answer:
9.265