The correct option is A 685.91 g
C+CO2→2CO (i)
Ni+4CO→Ni(CO)4 (ii)
In reaction (ii),
1 mol of Ni(CO)4 is produced from 4 mol of CO
170.7 g of Ni(CO)4 is produced from (4×28) g of CO
683 g of Ni(CO)4 will be obtained from =4×28170.7×683=448.13 g of CO
Yield of the reaction is given as 20 % Actual amount of CO required =448.130.2=2240.65 g
In (i) reaction,
2 mol of CO is produced by 1 mol of carbon
(2×28 g) of CO is produced by 12 g of carbon
2240.65 g will be produced by 122×28×2240.65=480.14 g
Since, the yield of the reaction is given as 70 %
Actual amount of carbon required =480.14/0.7=685.91 g