Limiting Reactants

Question

For the following reaction, 28.4 grams of carbon disulfide are allowed to react with 128 grams of chlorine gas .

carbon disulfide(s) + chlorine(g)  carbon tetrachloride(l) + sulfur dichloride(s)

What is the maximum amount of carbon tetrachloride that can be formed in grams?

What is the formula for the limiting reactant?

What amount of the excess reagent remains after the reaction is complete in grams?

Answer 1

First, you must calculate the moles present in each of the reactants by dividing the given masses of the reactants by their molar masses. We are using 3.61 moles of chlorine and 0.373 moles of carbon disulfide.

Then, we must write and balance the equation. Chlorine's coefficient is 8 and sulfur dichloride's is 2, so the reaction can be written as:

CS2 + 8Cl- --> CCl4 + 2SCl2

So for every mole of CS2 (carbon disulfide) there are 8 moles of chlorine, and for every mole of CS2, there is a mole of CCl4 (carbon tetrachloride).

If you multiply .373 by 8, it is around 2.9, less than our available chlorine moles. This means that because the ratio is larger than it should be, CS2 is the limiting reagent, which is consumed completely by the reaction, and chlorine is the excess reagent, which has some leftover.

That means there are 0.373 moles of CCl4 present and produced as well. Therefore, when multiplying the moles by the molar mass, the maximum amount of carbon tetrachloride (CCl4) is 57.375 grams, or .373 moles. The limiting reagent is CS2. 0.626 moles of the excess reagent, chlorine, are left, which is 22.193 grams.