\(4C + S_8 \rightarrow 4CS_2\)

\(\therefore \) Number of moles of carbon = \(\frac 4{12} = 0.33 \,mol\)

Number of moles of sulphur = \(\frac 8{256.8} = 0.031\, mol\)

\(\because \) 1 mole of S_{8} react with = 4 mole of carbon

\(\therefore \) \(\)0.031 mole of S_{8} react with = (4 x 0.031) mole of carbon

= 0.12 mol of carbon

but we have 0.33 moles of carbon

\(\therefore \) S_{8} is a limiting reagent.

\(\therefore \) Number of moles of CS_{2} formed = 0.031 x 4

\(\therefore \) Weight of CS_{2} = 0.031 x 4 x 76 = 9.42 g

Number of moles of carbon left = 0.33 - 0.12 = 0.21 mol

or

= 0.21 x 12 g

= 2.52 g

Hence, 9.42 g CS_{2} will formed and 2.52 g carbon left unreacted.