What mass of carbon disulphide, \(\mathrm{CS}_{2}\), can be completely oxidized to \(\mathrm{SO}_{2}\) and \(\mathrm{CO}_{2}\) by the oxygen liberated when \(325 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}_{2}\) react with water? (a) \(316.67 \mathrm{~g}\) (b) \(52.78 \mathrm{~g}\) (c) \(633.33 \mathrm{~g}\) (d) \(211.11 \mathrm{~g}\)

To start, let's write the balanced chemical equations for the reactions: For the decomposition of \(\mathrm{Na}_{2}\mathrm{O}_{2}\) in water: \(\mathrm{Na}_{2}\mathrm{O}_{2} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{NaOH} + \mathrm{H}_{2}\mathrm{O}_{2}\) For the decomposition of \(\mathrm{H}_{2}\mathrm{O}_{2}\) (since it is unstable and further decomposes): \(\mathrm{H}_{2}\mathrm{O}_{2} \rightarrow \mathrm{H}_{2}\mathrm{O} + \cfrac{1}{2}\mathrm{O}_{2}\) For the complete oxidation of \(\mathrm{CS}_{2}\): \(\mathrm{CS}_{2} + 3\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{SO}_{2}\) Now we can proceed to find the amount of oxygen produced.

First, find the moles of \(\mathrm{Na}_{2}\mathrm{O}_{2}\): given mass of \(\mathrm{Na}_{2}\mathrm{O}_{2} = 325 \mathrm{~g}\) molar mass of \(\mathrm{Na}_{2}\mathrm{O}_{2} = 2\times(23) + 2\times(16) = 78 \mathrm{~g/mol}\) So, moles of \(\mathrm{Na}_{2}\mathrm{O}_{2} = \cfrac{325 \mathrm{~g}}{78 \mathrm{~g/mol}} = 4.167 \mathrm{~mol}\) Now, we will use the stoichiometry of the first reaction to find the moles of \(\mathrm{H}_{2}\mathrm{O}_{2}\) produced. Number of moles of \(\mathrm{H}_{2}\mathrm{O}_{2} = \text{moles of }\mathrm{Na}_{2}\mathrm{O}_{2} = 4.167 \mathrm{~mol}\) Using the second reaction, we can now find the moles of oxygen produced: Number of moles of \(\mathrm{O}_{2} = 2\times \text{moles of }\mathrm{H}_{2}\mathrm{O}_{2} = 2\times 4.167 \mathrm{~mol} = 8.334 \mathrm{~mol}\)

Using the stoichiometry of the third reaction, we can find the moles of \(\mathrm{CS}_{2}\) that can be oxidized by the produced oxygen: Number of moles of \(\mathrm{CS}_{2} = \cfrac{\text{moles of }\mathrm{O}_{2}}{3} = \cfrac{8.334 \mathrm{~mol}}{3} = 2.778 \mathrm{~mol}\)

Finally, we will convert the moles of \(\mathrm{CS}_{2}\) to mass in grams: molar mass of \(\mathrm{CS}_{2} = 12 + 2\times(32) = 76 \mathrm{~g/mol}\) mass of \(\mathrm{CS}_{2} = \text{moles of }\mathrm{CS}_{2}\times \text{molar mass of }\mathrm{CS}_{2} = 2.778 \mathrm{~mol}\times 76 \mathrm{~g/mol} = 211.11 \mathrm{~g}\) Thus, the mass of carbon disulfide that can be completely oxidized is \(211.11 \mathrm{~g}\). The correct answer is (d) \(211.11 \mathrm{~g}\).