Sulphuric acid is produced when sulphur dioxide reacts with oxygen and water in the presence of a catalyst according to the reaction: ${2 S O}_{2} (g) + O_{2} (g) + {2 H}_{2} O (l) ⟶ {2 H}_{2} {S O}_{4}$ . If $5.6$ mol of $S O_{2}$ reacts with $4.8$ mol of $O_{2}$ and an excess of water, what is the maximum number of moles of $H_{2} {S O}_{4}$ that can be obtained?

2.5

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2.6

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5.4

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5.6

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Solution

The correct option is D $5.6$
The balanced reaction is given below:
$2 S O_{2} + O_{2} + 2 H_{2} O \to 2 H_{2} S O_{4}$
$5.6$ $4.8$ (No. of moles given)
$\frac{5.6}{2}$ $\frac{4.8}{1}$ (Ratio of moles given and stoichiometry coefficient )
= $2.8$ = $4.8$
As $2.8$ is lower than $4.8$ , $S O_{2}$ is the limiting reagent and product will form according to number of moles of $S O_{2}$ (given).
So, moles produced of $H_{2} S O_{4} = \frac{2}{2} \times 5.6 = 5.6$ mol.

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2 S O_{2} (g) + O_{2} (g) + 2 H_{2} O (l) \to 2 H_{2} S O_{4.}

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