Question

How many liters of ammonia would be produced from 40.0 grams of nitrogen?

Answer 1

At the condition of standard temperature and pressure called as STP, one mole of any gas contains the same amount of volume, known as 22.4 liters. This is called Avogadro’s law or hypothesis of gases.

Complete answer: The reaction of di- nitrogen, ${{N}_{2}}$with 3 moles of di- hydrogen, ${{H}_{2}}$, makes ammonia, $N{{H}_{3}}$as a product. The reaction is as follows:
${{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}$
As 1 mole of nitrogen gas produces 2 moles of ammonia, we have to calculate the number of moles in 40.0 g of nitrogen. The formula used will be,
Number of moles = $\dfrac{given\,mass}{molar\,mass}$
Number of moles in 40.0 g of ${{N}_{2}}$ = $\dfrac{given\,mass\,of\,{{N}_{2}}}{molar\,mass\,of\,{{N}_{2}}}$
Number of moles in 40.0 g of ${{N}_{2}}$ = $\dfrac{40.0\,g}{28\,g\,mo{{l}^{-1}}}$
Number of moles in 40.0 g of ${{N}_{2}}$ = 1.43 moles
We know, from Avogadro’s hypothesis, that 1 mole of a gas has 22.4 L volume at STP, so, 1 mole of nitrogen at STP will give, $2\times 22.4\,L\,of\,N{{H}_{3}}$ which is equal to 44.8 L of $N{{H}_{3}}$. Now, we will calculate the volume of $N{{H}_{3}}$ produced from 1.43 moles of ${{N}_{2}}$ by unitary method as,
Volume of $N{{H}_{3}}$= 1.43 mol of${{N}_{2}}$ $\times \dfrac{44.8\,L\,of\,N{{H}_{3}}}{1\,mol\,{{N}_{2}}}$
Volume of $N{{H}_{3}}$= 64.064 L
Hence, 40.0 gram of nitrogen will yield 64.064 L of ammonia.

Note: The calculation can also be done by taking out the number of moles produced from 1.43 moles of nitrogen and then by unitary method. The answer will be the same as 64.064 liters.