Question

According to the following reaction, how many grams of water are produced in the complete reaction of 29.7 grams of ammonia? 4 NH3(g) + 5 O2(g) + 4 NO(g) + 6 H2O(g) grams H2O

Answer 1

Answer: The mass of water produced in the reaction is 47.25 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of ammonia = 29.7 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{29.7g}{17g/mol}=1.75mol$

The given chemical reaction follows:

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

By stoichiometry of the reaction:

4 moles of ammonia produces 6 moles of water.

So, 1.75 moles of ammonia will produce = $\frac{6}{4}\times 1.75=2.625mol$ of water.

Now, calculating the mass of water by using equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 2.625 moles

Putting values in equation 1, we get:

$2.625mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=47.25g$

Hence, the mass of water produced in the reaction is 47.25 grams.