Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H 2 O that can be produced by combining 90.6 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Answer 1

Answer: The maximum mass of $H_2O$ produced is, 61.0 grams.

Explanation : Given,

Mass of $NH_3$ = 90.6 g

Mass of $O_2$ = 90.6 g

Molar mass of $NH_3$ = 17 g/mol

Molar mass of $O_2$ = 32 g/mol

First we have to calculate the moles of $NH_3$ and $O_2$.

$\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}$

$\text{Moles of }NH_3=\frac{90.6g}{17g/mol}=5.33mol$

and,

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}$

$\text{Moles of }O_2=\frac{90.6g}{32g/mol}=2.83mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

From the balanced reaction we conclude that

As, 5 mole of $O_2$ react with 4 mole of $NH_3$

So, 2.83 moles of $O_2$ react with $\frac{4}{5}\times 2.83=2.26$ moles of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 5 mole of $O_2$ react to give 6 mole of $H_2O$

So, 2.83 mole of $O_2$ react to give $\frac{6}{5}\times 2.83=3.39$ mole of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

Molar mass of $H_2O$ = 18 g/mole

$\text{ Mass of }H_2O=(3.39moles)\times (18g/mole)=61.0g$

Therefore, the maximum mass of $H_2O$ produced is, 61.0 grams.