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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` |
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Answer» 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` |
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