Question

To find the answer, we can actually write down the respective reaction which takes place. After that we can look out for what is given to us so that we can approach in a way that we can equate the reactant or product on equal ground with the same quantity.

Answer 1

Hint: To answer this question you should have the proper knowledge of the properties of hydrogen peroxide$\left( {{H_2}{O_2}} \right)$, its reaction with the compounds and elements given in the options above. You should also know the definition or meaning of the decomposition.

Complete answer:
\[N{a_2}C{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}O + C{O_2}\]
In question, it is said that weight of \[N{a_2}C{O_3}\] is \[6.2984g\] and after the product is obtained \[N{a_2}S{O_4}\] weighs \[8.4380g\].
We are aware that, for one molecule the number of moles are equal.
It can be any chemical formula, so for any given compound for one molecule the number of moles is equal at room temperature.
 So, we can actually equate those quantities, which are easily available to us to the one whose values are to be identified.
Here we have the value for \[mole = \dfrac{{given{\text{ }}mass}}{{total{\text{ atomic mass}}}}\]
So, for \[N{a_2}C{O_3}\], \[mole = \dfrac{{6.2984g}}{{46 + 12 + 16 \times 3}}\] this is equation (i)
Now, let us assume for sulphur the atomic mass is \[x\], then its mole formula will be:
\[mole = \dfrac{{8.4380g}}{{46 + x + 16 \times 4}}\] this is equation (ii).
We will equate (i) and (ii), we get
\[\dfrac{{6.2984g}}{{46 + 12 + 16 \times 3}} = \dfrac{{8.4380g}}{{46 + x + 16 \times 4}}\]
So, the mass of sulphur is 32g.

Note:
We should not forget to convert the equation SI units to the ones asked in the question. Suppose the quantity has SI units as kg, but if the question is asked in grams then you need to answer in grams only.