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)
It instead of given amount of sodium thiosulphate, 2 moles of sodium thiosulphate along with 3 moles of `CuSO_(4)` were taken initially . Then, moles of sodiu cuprothiosulphate formed is:

Options

A. 0
B. 1
C. 1.5
D. 2

Answer

A

Explanation

No moles of `Na_2S_2O_3` would remain in step (i) so further reaction will stop because `Na_2S_2O_3` is required is excess.