Question

2 moles of N2 is reacted with 4 moles of H2 to produce ammonia . Mass of ammonia produced is ??? ​

Answer 1

Answer:

32g

Explanation:

1 moles of N2 + 3 moles of H2 = 2 moles of NH3

Hence, We only have 4 moles of H. So H2 Becomes the Limiting Reagent.

So, Only one mole of Ammonia would be produced.

NH3 = 32g

Answer 2

Answer:

The mass of ammonia produced is 15.13 g.

Explanation:

$N_{2}(g) + 3H_{2}(g)$ → $2NH_{3}(g)$

⇒1 mole of nitrogen molecules reacts with 3 moles of hydrogen molecules to give 2 moles of ammonia gas.

⇒First, we need to find out the limiting reagent and excess reactant by dividing the given moles of each reactant by its coefficient.

⇒In the question, we have 2 moles of nitrogen molecules reacting with 4 moles of hydrogen molecules to produce a certain amount of ammonia gas.

$N_{2}$ → $\frac{2}{1} = 2$

$H_{2}$ → $\frac{4}{3} = 1.33$

⇒We can see that $N_{2}$ is the excess reactant while $H_{2}$ is the limiting reagent.

⇒Now we can calculate the amount of $NH_{3}$ produced.

$H_{2} : NH_{3}$

3     :    2

1.33 :    $x$

2 × 1.33 = 3 × $x$

2.66 = $3x$

0.89 mol = $x$

The molar mass of ammonia = 14 + 3 = 17

Mass of ammonia = 17 × 0.89 = 15.13 g