In the following reaction, how many grams of ammonia (NH3) will produce 300 grams of N2? 4NH3 + 6NO → 5N2 + 6H2O The molar mass of ammonia is 17.0337 grams and that of nitrogen is 28.02 grams.

Verified Answer :

145.899 g / 145.90 g

Answer: 145.87 grams of ammonia will produce the given amount of nitrogen.

Explanation:

To calculate the moles of nitrogen, we use the formula:

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

Given mass of nitrogen = 300 g

Molar mass of nitrogen = 28.02 g/mol

Putting values in above equation, we get:

\(\text{Moles of nitrogen}=\frac{300g}{28.02g/mol}=10.706mol\)

For the given reaction:

\(4NH_3+6NO\rightarrow 5N_2+6H_2O\)

By Stoichiometry of the reaction:

5 moles of nitrogen is produced from 4 moles of ammonia.

So, 10.706 moles of nitrogen will be produced from = \(\frac{4}{5}\times 10.706mol=8.564mol\)

Now, calculating the mass of ammonia, we use equation 1:

Molar mass of ammonia = 17.0337 g/mol

Moles of ammonia = 8.564 mol

Putting values in equation 1, we get:

\(8.564mol=\frac{\text{Mass of ammonia}}{17.0337g/mol}\\\\\text{Mass of ammonia}=145.87grams\)

Hence, 145.87 grams of ammonia will produce the given amount of nitrogen