632 g of sodium thiosulphate `(a_(2)S_(2)O_(3))` reacts with copper sulphate to form cuproc thiosulphate which is reduced by sodium thiosulphate to gi
632 g of sodium thiosulphate `(a_(2)S_(2)O_(3))` reacts with copper sulphate to form cuproc thiosulphate which is reduced by sodium thiosulphate to give cuprous compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothisosulphate `(Na_(4)[Cu_(6)(S_(2)O_(3))_(5)])`, (MW=1033)
`CuSO_(4) + Na_(2)S_(2)O_(3) rarr CuS_(2)O_(3) + Na_(2)SO_(4)" "` [very fast]
`2CuSO_(4)+Na_(2)S_(2)O_(3)+Na_(2)S_(2)O_(3) rarr Cu_(2)S_(2)O_(3) + Na_(2)S_(4)O_(6)`
`3Cu_(2)S_(2)O_(3) + 2Na_(2)S_(2)O_(3) rarr Na_(4)[Cu_(6)(S_(2)O_(3))_(5)]`
`" "` (Sodium cuprothisoulphate )
In this process , 0.2 mole of sodium cuprothiosulphate is formed .(O=16 , Na=23 , S=32)
Moles of sodium thiosulophate reacted and unreacted after the reaction are respectively,
A. `3 & 2`
B. `2 & 3`
C. `2.2 & 1.8`
D. `1.8 & 2.2`
1 Answers
Correct Answer - C
Moles of `Na_2S_2O_3` used =0.4+0.6+1.2=2.2
Moles of `Na_2S_2O_3` left unreacted =`632/158-2.2=1.8`