Question

1. calculate the number of grams of nh₃ produced by the reaction of 6.70 grams n₂

Explanation:

Step1: Write the balanced reaction

The reaction for the production of $\ce{NH3}$ from $\ce{N2}$ is the Haber's process: $\ce{N2 + 3H2 -> 2NH3}$ (assuming sufficient $\ce{H2}$ is present).

Step2: Calculate moles of $\ce{N2}$

Molar mass of $\ce{N2}$ is $2\times14 = 28\space g/mol$. Moles of $\ce{N2}$ = $\frac{\text{mass of } \ce{N2}}{\text{molar mass of } \ce{N2}}$ = $\frac{6.70\space g}{28\space g/mol}$ ≈ $0.2393\space mol$.

Step3: Use stoichiometry to find moles of $\ce{NH3}$

From the balanced equation, 1 mole of $\ce{N2}$ produces 2 moles of $\ce{NH3}$. So moles of $\ce{NH3}$ = $2\times$ moles of $\ce{N2}$ = $2\times0.2393\space mol$ = $0.4786\space mol$.

Step4: Calculate mass of $\ce{NH3}$

Molar mass of $\ce{NH3}$ is $14 + 3\times1 = 17\space g/mol$. Mass of $\ce{NH3}$ = moles $\times$ molar mass = $0.4786\space mol\times17\space g/mol$ ≈ $8.136\space g$.

Answer:

Approximately $\boldsymbol{8.14\space g}$ (or $8.13\space g$ depending on significant figures) of $\ce{NH3}$ is produced.