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Question: What weight of \(S{O_2}\)can be made by burning sulphur in \(5.0\) moles of oxygen? \(i)640{\text{...

What weight of SO2S{O_2}can be made by burning sulphur in 5.05.0 moles of oxygen?
i)640 gramsi)640{\text{ }}grams
ii)160 gramsii)160{\text{ }}grams
iii)80 gramsiii)80{\text{ }}grams
iv)320 gramsiv)320{\text{ }}grams

Explanation

Solution

In order to question this question, we need to first understand what a combustion reaction is and what will be the products formed when sulphur will burn in oxygen. Also, to find the weight that will be produced, we need to have the understanding of the mole concept and the balancing of the chemical reaction.

Complete answer:
In the given process where the sulphur will be burnt in the presence of a specific amount of oxygen, we first need to understand what will be the net chemical reaction of the above process.
The reaction will be: S(s)+O2(g)SO2(g)S(s) + {O_2}(g) \to S{O_2}(g)
Clearly, we can see one mole of sulphur reacts with one mole of oxygen molecule to give 1 mole of sulphur dioxide gas, i.e.SO2S{O_2}. hence, according to the given question, when 5.05.0moles of oxygen is used, we will get 5.05.0moles of sulphur burnt to give 5.05.0moles of sulphur dioxide, i.e.SO2S{O_2}.
Now, we need to convert these moles into amounts. Hence, we can easily use the molecular weight of the sulphur dioxide, i.e.SO2S{O_2} to give the answer.
Molecular weight of 1.01.0mole of SO2S{O_2} =32×1+16×2 = 32 \times 1 + 16 \times 2
Which gives the molecular weight of SO2S{O_2} =64grams = 64grams.
Hence, weight of 5.05.0moles of sulphur dioxide=5×64=320grams = 5 \times 64 = 320grams

Hence, the correct option is (iv)(iv).

Note:
Whenever solving these types of questions, make sure that the formed chemical reaction is always balanced, otherwise the calculation of number of moles will go wrong and which eventually will turn to wrong calculation of the amount of mass required to be found in the question. Also, make sure of differentiation between atomic mass and molecular mass and use whatever is required in the question.