Question

Calculate the moles of $S{{O}_{2}}$ produced by burning 1.0 metric ton (${{10}^{3}}$ kg) of coal containing 0.05% by mass of pyrites impurity.
(a) 8.32
(b) 4.16
(c) 12.48
(d) 2.08

Answer 1

Approach this question by calculating the mass of pyrite impurity in the coal by using mass percent as it is the way of expressing a concentration or describing the component in a particular mixture. Then using this calculate the number of moles of pyrite followed by the finding the moles of sulphur dioxide produced in the burning of coal.
Formula used:
We will require the following formulas:-
$\text{Mass percent =}\dfrac{\text{Mass of the solute}}{\text{Mass of the solution}}\times 100$
$n=\dfrac{\text{given mass}}{\text{Molar mass}}$

Complete answer:
According to the question, coal containing pyrite impurity ($Fe{{S}_{2}}$) is combusted in the presence of oxygen to produce $S{{O}_{2}}$.
-Calculation of mass of pyrite impurity in the coal by using mass percent as follows:-
The values provided in the question are:-
Mass percent of pyrite impurity ($Fe{{S}_{2}}$) = 0.05%
Here mass of solute is mass of pyrite impurity and mass of solution is mass of coal burnt.
Mass of coal burnt = ${{10}^{3}}$ kg = ${{10}^{6}}$ g
On substituting all the values in$\text{Mass percent =}\dfrac{\text{Mass of the solute}}{\text{Mass of the solution}}\times 100\%$, we get:-
$\begin{aligned} & \Rightarrow \text{0}.\text{05}\%\text{=}\dfrac{\text{Mass of the }Fe{{S}_{2}}}{\text{1}{{\text{0}}^{\text{6}}}}\times \text{100}\% \\\ & \Rightarrow \text{Mass of the }Fe{{S}_{2}}=\dfrac{\text{0}.\text{05}}{\text{100}}\times {{10}^{6}} \\\ & \Rightarrow \text{Mass of the }Fe{{S}_{2}}=500g \\\ \end{aligned}$
-Calculation of number of moles of $Fe{{S}_{2}}$:-
Molar mass of Fe = 56 g/mol
Molar mass of S = 32 g/mol
Molar mass of $Fe{{S}_{2}}$= (56 + 2(32)) g/mol = 120 g/mol
On substituting the values in$n=\dfrac{\text{given mass}}{\text{Molar mass}}$, we get:-
$n=\dfrac{500g}{120g/mol}=4.16moles$

-Calculation of moles of $S{{O}_{2}}$by using the chemical equation of burning of coal containing pyrite impurity as follows:-
$2Fe{{S}_{2}}+\dfrac{11}{2}{{O}_{2}}\to F{{e}_{2}}{{O}_{3}}+4S{{O}_{2}}$
As we can see that 2 moles of $Fe{{S}_{2}}$ produce 4 moles of$S{{O}_{2}}$, so 4.16 moles of $Fe{{S}_{2}}$will produce = $\text{4}\text{.17 moles of }Fe{{S}_{2}}\times \dfrac{4\text{ moles of S}{{\text{O}}_{2}}}{2\text{ moles of }Fe{{S}_{2}}}=8.32\text{ moles of S}{{\text{O}}_{2}}$
-Therefore, on 1.0 metric ton (${{10}^{3}}$ kg) of coal containing 0.05% by mass of pyrites impurity, (a) 8.32 moles of $S{{O}_{2}}$ is produced.

Note:
-Remember to convert the units of the given values in the question as per requirement so as to get an accurate answer and also understand the concept of concentration terms and moles in order to solve such types of questions.