Question

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

$A$: Calculate the mass of ammonia produced if $2.00\times {10}^{3}g$ dinitrogen reacts with $1.00\times {10}^{3}g$ of dihydrogen.

$B$: Will any of the two reactants remain unreacted?

$C$: If yes, which one and what would be its mass?

Answer 1

Balance the equation

Balancing the given chemical equation,
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$


Given amount of $N_2=2.00\times {10}^{3}g=2000g\text{of}{N}_2$ 
Given amount of $H_2=1.00\times {10}^{3}g=1000g\text{of}{H}_2$

Amount of $H_2$ required
From the equation, $1$ mol (28g) of $N_2$ reacts with $3$ mol(6g) of $H_2$ to give $2$ mol (34g) of $N{H}_2$

Mass of $H_2$ that means with $2000g\text{of}{N}_2=\frac{6}{28}\times 2000=428.6g\text{of}{H}_2$
Here, $N_2$ is consumed completely, hence it is the limiting reagent.
Hence, some amount of $H_2$ will remain unreacted.

Amount of $N{H}_3$ produced

Now, $28g$ of $N_2$ produces $34g\text{of}N{H}_3$

Hence,
$\text{mass of}N{H}_3\text{produced by}2000g\text{of}{N}_2=\frac{34}{28}\times 2000gN{H}_3$
$=2428.57gN{H}_3$

Final Answer: So the mass of ammonia produced is $2428.57g$

$\left(ii\right)$ Here, $N_2$ is the limiting reagent and $H_2$ is the excess reagent. Hence, $H_2$ will remain unreacted.

$\left(iii\right)$ Mass of dihydrogen left unreacted $=(1\times {10}^3-428.6)g$
$571.4g$