Question

what is the greatest amount of NH3 moles that can be produced from 3.2 moles of N2 and 5.4 moles of H2 which is a limiting reagent and excess reactant

Answer 1

Explanation:

N2 + 3H2 ——> 2NH3

1 Mole n2 requires 3 moles of h2

3.2 mole req 3 ×3.2 which is equal to 9.6 mole

but we are given 5.4 mole h2 only

:. therefore , limiting reagent is h2 and excess reagent is n2

now 3 mole h2 = 2 mole nh3

5.4 mole will have nh3 = 2 ×5.4/ 3 = 3.6 mole

Answer 2

Answer:

Limiting reagent - $H_{2}$

Excess reactant - $N_{2}$

Total moles of $NH_{3}$ produced - 3.6 mol

Explanation:

$N_{2} (g) + 3H_{2} (g)$ → $2NH_{3} (g)$

First divide the given moles of each reactant with its coefficient to find out the limiting reagent and excess reactant.

$N_{2}$ → $\frac{3.2}{1} = 3.2$

$H_{2}$ → $\frac{5.4}{3} = 1.8$

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

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

$H_{2} : NH_{3}$

3     :      2

5.4  :      $x$

3 × $x$ = 2 × 5.4

3$x$ = 10.8

$x = \frac{10.8}{3}$

$x$ = 3.6 moles of $NH_{3}$