Question

632 g of sodium thiosuphate $\left(N{a}_2{S}_2{O}_3\right)$ reacts with copper sulphate to form cupric thiosulphate which is reduced by sodium thiosulphate to given cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothiosulphate $N{a}_4\left[C{u}_5\right(S_2{O}_3{)}_5\left]\right)$ $CuS{O}_4+N{a}_2{S}_2{O}_3\to Cu{S}_2{O}_3+N{a}_{2}S{O}_4$
$2Cu{S}_2{O}_3+N{a}_2{S}_2{O}_3\to C{u}_2{S}_2{O}_3+N{a}_2{S}_4{O}_6$
$3Cu{S}_2{O}_3+2N{a}_2{S}_2{O}_3\to N{a}_4\left[C{u}_6\right(S_2{O}_3{)}_5]$
In this process 0.2 mole of sodium cuprothiosulphate is formed. (O=16, Na=23, S=32)
If instead of given amount of sodium thiosulphate, 2 moles of sodium thiosulphate along with 3 moles of $CuS{O}_4$ were taken initially. Then moles of sodium cuprothiosulphate formed is:

• A. 0
• B. 1
• C. 1.5
• D. 2

Answer 1

The correct option is A 0

No moles of $\left(N{a}_2{S}_2{O}_3\right)$ would remain in step(i) so further reaction will stop because $\left(N{a}_2{S}_2{O}_3\right)$ is required in excess.