A colourless gas A has a choking smell. It turns acidified potassium dichromate from orange to…

Step 1. The compounds. A is $\text{SO}_{2}$: a choking smell, and a reducing gas. With excess alkali it gives B, sodium sulphite: $\text{SO}_{2}\ +\ 2\text{NaOH}\ \rightarrow \ \text{Na}_{2}\text{SO}_{3}\ +\ \text{H}_{2}\text{O}$. Boiling with sulphur gives C, sodium thiosulphate: $\text{Na}_{2}\text{SO}_{3}\ +\ \text{S}\ \rightarrow \ \text{Na}_{2}\text{S}_{2}\text{O}_{3}$. With iodine it gives D, sodium tetrathionate: $2\text{Na}_{2}\text{S}_{2}\text{O}_{3}\ +\ \text{I}_{2}\ \rightarrow \ \text{Na}_{2}\text{S}_{4}\text{O}_{6}\ +\ 2\text{NaI}$.

Step 2. The states. In $\text{Na}_{2}\text{S}_{2}\text{O}_{3}$: $2(+1) + 2x + 3(-2) = 0$, so $x = +2$. In $\text{Na}_{2}\text{S}_{4}\text{O}_{6}$: $2(+1) + 4x + 6(-2) = 0$, so $x = +2.5$.

Step 3. Moles. $\dfrac{6.40\ \text{g}}{64.0\ \text{g/mol}} = 0.100\ \text{mol}$ of $\text{SO}_{2}$. One mole gives one mole of B, and one mole of C.

Step 4. Mass of C. $0.100\ \text{mol} \times 158.0\ \text{g/mol} = 15.8\ \text{g}$.

Step 5. The iodine. Two thiosulphate for each iodine: $0.0500\ \text{mol}$, which is $0.0500 \times 253.8 = 12.7\ \text{g}$.