Question

Carbon disulphide, $C{S}_2,$ can be made from by-product $S{O}_2.$ The overall reaction is $5C+2S{O}_2\to C{S}_2+4CO$
How much $C{S}_2$ in kg can be produced from 440 kg of waste $S{O}_2$ with 60 kg of coke if the $S{O}_2$ conversion is 80%

Answer 1

Purity of $S{O}_2$ = 80 %
Mass of pure $S{O}_2$ = $440\times 0.8=352\text{kg}$
$5C+2S{O}_2\to C{S}_2+4CO$
Moles of $S{O}_2=\frac{352\times {10}^3\text{g}}{64\text{g/mol}}=5.5\times {10}^3\text{mol}$
Moles of $C=\frac{60\times {10}^3\text{g}}{12\text{g/mol}}=5\times {10}^3\text{mol}$

$\frac{\text{moles of}S{O}_2}{\text{stoichiometric coefficient}}=\frac{5.5\times {10}^3}{2}$
$\frac{\text{moles of}C}{\text{stoichiometric coefficient}}=\frac{5\times {10}^3}{5}$

Hence C is the limiting reagent.
1 Mole of carbon disulphide formed from 5 moles of C.
Hence moles of carbon disulphide formed from $5\times {10}^3$ = $\frac{1}{5}\times 5\times {10}^3=1000$ mol
mass of $C{S}_2$ $=1000\times 76g=76kg$