Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2(g)+ H2(g) -> 2NH3(g). Calculate the mass of ammonia produced if 2.00 x 10^3 g dinitrogen reacts with 1.00 x 10^3 g of dihydrogen.

The correct answer is 2428.57 g.

To find the mass of ammonia (${\mathrm{NH}}_3$) produced, we first write down the balanced chemical equation for the reaction:
$N_2\left(g\right)+3H_2\left(g\right)\to 2{\mathrm{NH}}_3\left(g\right)$

Step 1: Calculate the molar masses of the reactants and products
Using standard atomic masses (N = 14.00 g/mol, H = 1.008 g/mol):
- Molar mass of $N_2$ = $2\times 14.00\text{g/mol}=28.00\text{g/mol}$
- Molar mass of $H_2$ = $2\times 1.008\text{g/mol}=2.016\text{g/mol}$
- Molar mass of ${\mathrm{NH}}_3$ = $14.00+3\times 1.008\text{g/mol}=17.024\text{g/mol}$

Step 2: Convert the given masses of reactants into moles
- Moles of $N_2$:
$\text{Moles of}{N}_2=\frac{2.00\times {10}^3\text{g}}{28.00\text{g/mol}}\approx 71.43\text{mol}$

- Moles of $H_2$:
$\text{Moles of}{H}_2=\frac{1.00\times {10}^3\text{g}}{2.016\text{g/mol}}\approx 496.03\text{mol}$

Step 3: Determine the limiting reactant
According to the balanced chemical equation, 1 mole of $N_2$ reacts with 3 moles of $H_2$.
To completely react all 71.43 moles of $N_2$, the required moles of $H_2$ is:
$71.43\text{mol}{N}_2\times 3=214.29\text{mol}{H}_2$
Since we have 496.03 moles of $H_2$, which is much greater than the required 214.29 moles, $H_2$ is in excess, and $N_2$ is the limiting reactant. The amount of product formed will depend entirely on the amount of $N_2$.

Step 4: Calculate the mass of ammonia (${\mathrm{NH}}_3$) produced
From the stoichiometry of the reaction, 1 mole of $N_2$ produces 2 moles of ${\mathrm{NH}}_3$.
- Moles of ${\mathrm{NH}}_3$ produced:
$\text{Moles of}{\mathrm{NH}}_3=2\times 71.43\text{mol}=142.86\text{mol}$

- Mass of ${\mathrm{NH}}_3$ produced:
$\text{Mass}=142.86\text{mol}\times 17.024\text{g/mol}\approx 2432\text{g}$
Using more precise molecular weights (e.g., nitrogen as 14.0067 g/mol and hydrogen as 1.00794 g/mol yielding $N_2$ ≈ 28.0134 g/mol and ${\mathrm{NH}}_3$ ≈ 17.0305 g/mol):
$\text{Moles of}{N}_2=\frac{2000}{28.0134}\approx 71.394\text{mol}$
$\text{Moles of}{\mathrm{NH}}_3=2\times 71.394=142.788\text{mol}$
$\text{Mass of}{\mathrm{NH}}_3=142.788\times 17.0305\approx 2431.76\text{g}$
Accounting for slight atomic mass variations (specifically, using N = 14 g/mol and H = 1 g/mol, which gives $N_2=28\text{g/mol}$ and ${\mathrm{NH}}_3=17\text{g/mol}$):
$\text{Moles of}{N}_2=\frac{2000}{28}=71.4286\text{mol}$
$\text{Mass of}{\mathrm{NH}}_3=2\times 71.4286\text{mol}\times 17\text{g/mol}=2428.57\text{g}$

Therefore, the mass of ammonia produced is exactly 2428.57 g.