Question

When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

Answer 1

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

Mass of $H_2O$ = 75.0 g

Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$.

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.