If you mix 10.0 g of carbon with 20.0 g of sulfur, S8, and the reaction 4C+S8→4CS2 occurs, what is the limiting reactant?

Answer 1

We can see from the balanced equation that we need 4 times as much carbon as sulfur.

First convert both reactants to moles

C: 10 g C x (1 mol C / 12 g C) = 0.83 mol C

S₈: 20 g S₈ x (1 mol S₈ / 256 g S₈) = 0.078 mol S₈

Take the ratio of C to S₈

0.83 mol C / 0.078 mol S₈ = 10.7

This tells us that we have 10.7 times as much carbon as S₈.

Thus, in this case, S₈ is the limiting reagent.

Alternatively, we can see which reactant produces the least amount of product and that reactant will be the limiting reactant.

For this problem:

0.83 mol C x (4 mol CS₂ / 4 mol C) = 0.83 mol CS₂

0.078 mol S₈ (4 mol CS₂ / 1 mol S₈) = 0.312 mol CS₂

Comparing, we can see that 0.078 mol of S₈ produces less CS₂ than 0.83 mol C. Thus, S₈ is the limiting reactant.