Question

The energy required for the chemical reaction of 21.7 grams of nitrogen to produce ammonia is how many kilojoules? If you put something that isn't the answer I will report you.

Answer 1

Answer:

$Q=-71.1kJ$

Explanation:

Hello!

In this case, since the formation of ammonia by starting with nitrogen and therefore hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Which has an energy of reaction of:

$\Delta _fH_{NH_3}=-45.90 \frac{kJ}{molNH_3}$

We can compute the energy required for this reaction by first computing the moles of ammonia yielded by 21.7 grams of nitrogen (28.02 g/mol) via stoichiometry:

$n_{NH_3}=21.7gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=1.55molNH_3$

Thus, the energy turns out:

$Q=n_{NH_3}\Delta _fH_{NH_3}=1.55molNH_3 * -45.90 \frac{kJ}{molNH_3}\\\\Q=-71.1kJ$

Best regards!