Question

Number of moles of sodium thiosulphate reacted and unreacted after the reaction respectively is :

• A. 3 and 2
• B. 2 and 3
• C. 2.2 and 1.8
• D. 1.8 and 2.2

Answer 1

Answer: (C)

2.2 and 1.8

$CuS{O}_4+\underset{1.2mole}{N{a}_2{S}_2{O}_3}\to \underset{1.2mole}{Cu{S}_2{O}_3}+N{a}_{2}S{O}_4$

$\underset{1.2mole}{2Cu{S}_2{O}_3}+\underset{0.6mole}{N{a}_2{S}_2{O}_3}\to \underset{0.6mole}{C{u}_2{S}_2{O}_3}+N{a}_2{S}_4{O}_6$

$\underset{0.6mole}{3C{u}_2{S}_2{O}_3}+\underset{0.4mole}{2N{a}_2{S}_2{O}_3}\to \underset{0.2mole}{N{a}_4\left(C{u}_6\right(S_2{O}_3{)}_5)}$

Moles of $N{a}_2{S}_2{O}_3$ used $=0.4+0.6+1.2=2.2$.

Total number of moles of $N{a}_2{S}_2{O}_3$ present $=\frac{632}{158}$.

Moles of $N{a}_2{S}_2{O}_3$ left unreacted $=\frac{632}{158}-2.2=1.8$.