Question

Sulphuric acid is produced when sulphur dioxide reacts with oxygen and water in the of a catalyst according to the reaction: $2S{{O}_{2}}(g)+{{O}_{2}}(g)+2{{H}_{2}}O(l)\to 2{{H}_{2}}S{{O}_{4}}$. 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?
A) 2.5
B) 2.6
C) 5.4
D) 5.6

Answer 1

The answer to this question is based on writing the balanced equation given and then calculating the stoichiometric coefficient and then followed by finding the limiting reagent.

Complete answer:
In the lower classes of chemistry, we have studied about the stoichiometry and the stoichiometric coefficients and also about the definition of the limiting reagent.
Let us now see the calculation of the maximum number of moles of sulphuric acid that is obtained.
Firstly, we shall write the balanced chemical equation as per the data given which is as follows,
$2S{{O}_{2}}(g)+{{O}_{2}}(g)+2{{H}_{2}}O(l)\to 2{{H}_{2}}S{{O}_{4}}$
Here, the number of moles of sulphur dioxide that is the reactant is given as 5.6 mol and that of oxygen which is also a reactant is given as 4.8 mol.
Now, the ratio of these values as per the number of atoms present in the reactant side will be, $\dfrac{5.6}{2}and\dfrac{4.8}{1}$ respectively. Thus, the stoichiometric coefficient for these by simplification can be written as 2.8 and 4.8.
Here, since the stoichiometric coefficient of sulphur dioxide is low, this will be the limiting reagent and therefore the number of moles of sulphuric acid obtained can be calculated based on the number of moles of sulphur dioxide formed as,
$\Rightarrow \dfrac{2}{2}\times 5.6=5.6mol$

Therefore, the correct answer is option D) 5.6.

Note:
Note that the reaction of sulphur dioxide with oxygen and water in air forms sulphuric acid and this process is nothing but the formation of acid rain. This fact helps you to approach the correct equation if the data is given in the form of words as ‘acid rain’.