Question

1. how many grams of nh₃ are produced when 89.1 grams of n₂ gas react according to the following balanced chemical equation. show your work for full credit. n₂ + 3h₂ → 2nh₃

Explanation:

Step1: Determine molar masses

The molar mass of $N_2$ is $M_{N_2}=2\times14\ g/mol = 28\ g/mol$, and the molar mass of $NH_3$ is $M_{NH_3}=14 + 3\times1=17\ g/mol$.

Step2: Calculate moles of $N_2$

The number of moles of $N_2$, $n_{N_2}=\frac{m_{N_2}}{M_{N_2}}$. Given $m_{N_2} = 801\ g$, so $n_{N_2}=\frac{801\ g}{28\ g/mol}\approx28.61\ mol$.

Step3: Use mole - ratio from balanced equation

The balanced chemical equation is $N_2+3H_2
ightarrow2NH_3$. The mole - ratio of $N_2$ to $NH_3$ is $1:2$. So for every 1 mole of $N_2$ that reacts, 2 moles of $NH_3$ are produced.

Step4: Calculate moles of $NH_3$

$n_{NH_3}=2\times n_{N_2}$. Substituting $n_{N_2}\approx28.61\ mol$, we get $n_{NH_3}=2\times28.61\ mol = 57.22\ mol$.

Step5: Calculate mass of $NH_3$

The mass of $NH_3$, $m_{NH_3}=n_{NH_3}\times M_{NH_3}$. Substituting $n_{NH_3}=57.22\ mol$ and $M_{NH_3}=17\ g/mol$, we have $m_{NH_3}=57.22\ mol\times17\ g/mol = 972.74\ g$.

Answer:

972.74 g