Question

What mass of carbon disulphide, $C{{S}_{2}}$ can be completely oxidized to $S{{O}_{2}}$ and $C{{O}_{2}}$ by the oxygen liberated when 325 g of $N{{a}_{2}}{{O}_{2}}$ react with water?
A) 316.67 g
B) 52.78 g
C) 633.33 g
D) 211.11 g

Answer 1

Hint: The Answer to this question lies in the concept of physical chemistry which is based on the calculation of number of moles of $N{{a}_{2}}{{O}_{2}}$ given and then calculating the number of moles of the reactants used to produce the product and then the mass of $C{{S}_{2}}$ which can be completely oxidised.

Complete Solution :
In the lower classes of chemistry, we have come across the concepts of physical chemistry which deals with the basic concepts of calculation of number of moles present in the given amount of the compound and related entities.
- Now, let us calculate the required answer in the similar method.
From the data given, we have.
325 g of $N{{a}_{2}}{{O}_{2}}$ is being reacted with water and now we have to find the number of moles of this compound. This is as follows,
Molar mass of$N{{a}_{2}}{{O}_{2}}$ is $2\times 23+2\times 16 = 78~g/mol$
Therefore, number of moles of$N{{a}_{2}}{{O}_{2}}$given will be$ = \dfrac{325}{78} = 4.166~moles$
- Now, let us write the reaction that is taking place when $N{{a}_{2}}{{O}_{2}}$ react with water.
\[2N{{a}_{2}}{{O}_{2}}+2{{H}_{2}}O\to 4NaOH+{{O}_{2}}\]
Thus, by this equation we can say that 2 moles of $N{{a}_{2}}{{O}_{2}}$ gives 1 mole of ${{O}_{2}}$
Hence, 4.166 moles of$N{{a}_{2}}{{O}_{2}}$ gives $ =\dfrac{4.166}{2} = 2.08~ moles$ of ${{O}_{2}}$

- Now, let us calculate the number of moles of $C{{S}_{2}}$ required to react with oxygen which is based on the balanced equation shown below,
\[C{{S}_{2}}+3{{O}_{2}}\to C{{O}_{2}}+2S{{O}_{2}}\]
Now, the molar mass of $C{{S}_{2}}$ is 76 g/mol.
From the above chemical equation, we can say that 1 mole of $C{{S}_{2}}$ needs 3 moles of ${{O}_{2}}$
Thus, 2.083 moles of ${{O}_{2}}$ requires let us say ‘x’ moles of $C{{S}_{2}}$ to produce $S{{O}_{2}}$ and $C{{O}_{2}}$
Therefore, by simple cross multiplication rule,
\[x=\dfrac{2.083\times 1}{3}=0.6943~moles\] of $C{{S}_{2}}$
.Therefore, by these data we can calculate the mass of $C{{S}_{2}}$ that can be completely oxidised by 2.083 moles of ${{O}_{2}}$ and that is $ = 0.6943\times 76 = 52.76~g$
So, the correct answer is “Option B”.

Note: Important point to note is that the oxygen liberated by the reaction of $N{{a}_{2}}{{O}_{2}}$ with water is used for the oxidation of $C{{S}_{2}}$ and do not confuse this with the balanced equation as to which forms reactants and which forms the product.