Question

Calculate the moles of ${\text{S}}{{\text{O}}_{\text{2}}}$ produced by burning ${\text{1}}{\text{.0}}$ metric ton $\left( {{{10}^3}{\text{ kg}}} \right)$ of coal containing $0.05\%$ by mass of pyrites impurity.
(A) 8.32
(B) 4.16
(C) 12.48
(D) 2.08

Answer 1

To solve this we must know the molecular formula of pyrite. The molecular formula of pyrite is ${\text{Fe}}{{\text{S}}_{\text{2}}}$. From the given percent by mass calculate the mass of pyrite in ${\text{1}}{\text{.0}}$ metric ton $\left( {{{10}^3}{\text{ kg}}} \right)$ of coal. Then calculate the number of moles of pyrite. Then by writing the balanced chemical equation, calculate the moles of ${\text{S}}{{\text{O}}_{\text{2}}}$ produced.

Formula Used:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$

Complete step-by-step solution: We know that pyrite is a mineral of sulphide. It is also known as iron sulphide. The molecular formula of pyrite i.e. iron sulphide is ${\text{Fe}}{{\text{S}}_{\text{2}}}$.
We are given that ${\text{1}}{\text{.0}}$ metric ton $\left( {{{10}^3}{\text{ kg}}} \right)$ of coal contains $0.05\%$ by mass of pyrites impurity. Thus,
Mass of pyrite in ${10^3}{\text{ kg}}$ coal $= 1000{\text{ kg}} \times \dfrac{{0.05}}{{100}}$
Mass of pyrite in ${10^3}{\text{ kg}}$ coal $= 0.5{\text{ kg}} = 500{\text{ g}}$
Calculate the number of moles of ${\text{Fe}}{{\text{S}}_{\text{2}}}$ in $500{\text{ g}}$ of ${\text{Fe}}{{\text{S}}_{\text{2}}}$ using the equation as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $500{\text{ g}}$ for the mass of ${\text{Fe}}{{\text{S}}_{\text{2}}}$, $120{\text{ g/mol}}$ for the molar mass of ${\text{Fe}}{{\text{S}}_{\text{2}}}$. Thus,
${\text{Number of moles of Fe}}{{\text{S}}_2} = \dfrac{{500{\text{ g}}}}{{120{\text{ g/mol}}}}$
${\text{Number of moles of Fe}}{{\text{S}}_2} = 4.17{\text{ mol}}$
Thus, the number of moles of ${\text{Fe}}{{\text{S}}_{\text{2}}}$ are $4.17{\text{ mol}}$.
The balanced chemical equation when pyrite is burned in presence of oxygen is as follows:
${\text{2Fe}}{{\text{S}}_{\text{2}}} + {\text{5}}{{\text{O}}_{\text{2}}} \to {\text{4S}}{{\text{O}}_{\text{2}}} + {\text{2FeO}}$
From the reaction stoichiometry,
$1{\text{ mol Fe}}{{\text{S}}_2} = 2{\text{ mol S}}{{\text{O}}_2}$
Thus, $4.17{\text{ mol}}$ of ${\text{Fe}}{{\text{S}}_{\text{2}}}$ will produce,
${\text{Number of moles of S}}{{\text{O}}_2} = 4.17{\text{ mol Fe}}{{\text{S}}_2} \times \dfrac{{2{\text{ mol S}}{{\text{O}}_2}}}{{{\text{1 mol Fe}}{{\text{S}}_2}}}$
${\text{Number of moles of S}}{{\text{O}}_2} = 8.32{\text{ mol}}$
Thus, the moles of ${\text{S}}{{\text{O}}_{\text{2}}}$ produced by burning ${\text{1}}{\text{.0}}$ metric ton $\left( {{{10}^3}{\text{ kg}}} \right)$ of coal containing $0.05\%$ by mass of pyrites impurity are $8.32{\text{ mol}}$.

Thus, the correct option is (A) 8.32.

Note: Remember to write the correct balanced equation for the reaction in which pyrite is burned in presence of oxygen. Incorrect balanced chemical equations will lead to incorrect stoichiometry which leads to incorrect answers.