Question

Consider the following equation for the combustion of ammonia. Assume all substances are gases measured under the same conditions. 4NH3 + 3O2 → 2N2 + 6H2O How many liters of N2 is formed when 125 L of NH3 is burned?

Answer 1

Answer:

Therefore 62.5 L of N₂ is formed when 125 L of NH₃ is burned.

Explanation:

Given,

4 NH₃+3O₂→2N₂+6H₂O

The coefficient of balanced equation indicates the mole of the reactant  and the products.

The volume ratio is the same as the mole ratio when volume is measured under sane condition.

Therefore,

$\frac{\textrm{volume of }NH_3}{\textrm{volume of }N_2} =\frac{4}{2}$

Given that the volume of NH₃ is 125 L

$\frac{125L}{\textrm{volume of }N_2} =\frac{4}{2}$

$\Rightarrow {\textrm{volume of }N_2} =\frac{2\times 125L}{4}$

$\Rightarrow {\textrm{volume of }N_2} =62.5 L$

Therefore 62.5 L of N₂ is formed when 125 L of NH₃ is burned.