Question

If $340{\text{ g}}$ of a mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ in the incorrect ratio gave $20\%$ yield of ${\text{N}}{{\text{H}}_{\text{3}}}$, the mass produced will be:
A. $16{\text{ g}}$
B. $17{\text{ g}}$
C. $20{\text{ g}}$
D. $68{\text{ g}}$

Answer 1

Follow the given steps to calculate the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ and the mass of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ is $340{\text{ g}}$:
Write the balanced chemical equation.
Determine the molar masses of ${{\text{N}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$.
Determine the masses of ${{\text{N}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ for $100\%$ yield.
Determine the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$.
Determine the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ and the mass of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ is $340{\text{ g}}$.

Complete step by step answer:
Write the balanced chemical equation as follows:
In the reaction, ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ react to produce ${\text{N}}{{\text{H}}_{\text{3}}}$. Thus, the chemical equation is,
${{\text{N}}_{\text{2}}} + {{\text{H}}_{\text{2}}} \to {\text{N}}{{\text{H}}_{\text{3}}}$
Count the number of each atom on the reactant and product side. Thus,

$${{\text{N}}_{\text{2}}} + {{\text{H}}_{\text{2}}} \to {\text{N}}{{\text{H}}_{\text{3}}} \\\ {\text{ N}} - 2{\text{ N}} - 1 \\\ {\text{ H}} - 2{\text{ H}} - 3 \\\$$

Change the coefficient of ${\text{N}}{{\text{H}}_{\text{3}}}$ to $2$ to balance the ${\text{N}}$ atoms. Thus,

$${{\text{N}}_{\text{2}}} + {{\text{H}}_{\text{2}}} \to 2{\text{N}}{{\text{H}}_{\text{3}}} \\\ {\text{ N}} - 2{\text{ N}} - 2 \\\ {\text{ H}} - 2{\text{ H}} - 6 \\\$$

Change the coefficient of ${{\text{H}}_{\text{2}}}$ to $3$ to balance the ${\text{H}}$ atoms. Thus,

$${{\text{N}}_{\text{2}}} + 3{{\text{H}}_{\text{2}}} \to 2{\text{N}}{{\text{H}}_{\text{3}}} \\\ {\text{ N}} - 2{\text{ N}} - 2 \\\ {\text{ H}} - 6{\text{ H}} - 6 \\\$$

The number of all the atoms on the reactant and product side are equal. Thus, the balanced chemical equation is,
${{\text{N}}_{\text{2}}} + 3{{\text{H}}_{\text{2}}} \to 2{\text{N}}{{\text{H}}_{\text{3}}}$
Step 2: Determine the molar masses of ${{\text{N}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ as follows:
Calculate the molar mass of ${{\text{N}}_{\text{2}}}$ as follows:
${\text{Molar mass of }}{{\text{N}}_{\text{2}}} = 2 \times {\text{Mass of N}} \\\ = 2 \times 14 \\\ {\text{Molar mass of }}{{\text{N}}_{\text{2}}} = 28{\text{ g mo}}{{\text{l}}^{ - 1}} \\\$
Thus, the molar mass of ${{\text{N}}_{\text{2}}}$ is $28{\text{ g mo}}{{\text{l}}^{ - 1}}$.
Calculate the molar mass of ${{\text{H}}_{\text{2}}}$ as follows:
${\text{Molar mass of }}{{\text{H}}_{\text{2}}} = 2 \times {\text{Mass of H}} \\\ = 2 \times 1 \\\ {\text{Molar mass of }}{{\text{H}}_{\text{2}}} = 2{\text{ g mo}}{{\text{l}}^{ - 1}} \\\$
Thus, the molar mass of ${{\text{H}}_{\text{2}}}$ is $2{\text{ g mo}}{{\text{l}}^{ - 1}}$.
Calculate the molar mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ as follows:
${\text{Molar mass of N}}{{\text{H}}_{\text{3}}} = \left( {1 \times {\text{Mass of N}}} \right) + \left( {3 \times {\text{Mass of H}}} \right) \\\ = \left( {1 \times 14} \right) + \left( {3 \times 1} \right) \\\ = 14 + 3 \\\ {\text{Molar mass of N}}{{\text{H}}_{\text{3}}} = 17{\text{ g mo}}{{\text{l}}^{ - 1}} \\\$
Thus, the molar mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ is $17{\text{ g mo}}{{\text{l}}^{ - 1}}$.
Step 3: Determine the masses of ${{\text{N}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ as follows:
From, the reaction stoichiometry, $1{\text{ mol }}{{\text{N}}_2}$ reacts with ${\text{3 mol }}{{\text{H}}_{\text{2}}}$ to produce ${\text{2 mol N}}{{\text{H}}_{\text{3}}}$.
Thus,
$1{\text{ mol }}{{\text{N}}_2} = 1 \times 28 = 28{\text{ g}}$
${\text{3 mol }}{{\text{H}}_{\text{2}}} = 3 \times 2 = 6{\text{ g}}$
${\text{2 mol N}}{{\text{H}}_{\text{3}}} = 2 \times 17 = 34{\text{ g}}$
Thus, $28{\text{ g}}$ of ${{\text{N}}_{\text{2}}}$ reacts with $6{\text{ g}}$ of ${{\text{H}}_{\text{2}}}$ to produce $34{\text{ g}}$ of ${\text{N}}{{\text{H}}_{\text{3}}}$.
Mass of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ mixture $= 28 + 6 = 34{\text{ g}}$.
Thus, $34{\text{ g}}$ of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ reacts to produce $34{\text{ g}}$ of ${\text{N}}{{\text{H}}_{\text{3}}}$. This id $100\%$ yield.
Step 4: Determine the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ as follows:
$100\%$ yield produces $34{\text{ g}}$ of ${\text{N}}{{\text{H}}_{\text{3}}}$. Thus, the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ is,
${\text{Mass of N}}{{\text{H}}_3} = 34{\text{ g}} \times \dfrac{{20}}{{100}} \\\ {\text{Mass of N}}{{\text{H}}_3} = 6.8{\text{ g}} \\\$
Thus, the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ is $6.8{\text{ g}}$.
Step 5: Determine the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ and the mass of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ is $340{\text{ g}}$ as follows:
When the reaction yield is $20\%$, $34{\text{ g}}$ of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ produces $6.8{\text{ g}}$ of ${\text{N}}{{\text{H}}_{\text{3}}}$. Thus, when the reaction yield is $20\%$, $340{\text{ g}}$ of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ produces,
${\text{Mass of N}}{{\text{H}}_3} = 6.8{\text{ g N}}{{\text{H}}_3} \times \dfrac{{340{\text{ }}\not{{{\text{g mixture of }}{{\text{N}}_2}{\text{ and }}{{\text{H}}_2}}}}}{{34{\text{ }}\not{{{\text{g mixture of }}{{\text{N}}_2}{\text{ and }}{{\text{H}}_2}}}}} \\\ {\text{Mass of N}}{{\text{H}}_3} = 68{\text{ g}} \\\$
Thus, the mass of ${\text{N}}{{\text{H}}_{\text{3}}}$ produced when the yield of reaction is $20\%$ and the mass of mixture of ${{\text{N}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}$ is $340{\text{ g}}$ is $68{\text{ g}}$.

So, the correct answer is Option D.

Note:
Write the correct balanced chemical equation for the given reaction. Unbalanced chemical equations can lead to incurred masses of the reactants and products.