Dr. DJ W. answered • 08/11/20

Highly Experienced Chemistry Tutor|UCLA PhD Scientist|10+ yrs|

# First you need the balanced Equation as below:

# CS_{2} (s) + 4 Cl_{2} (g) → CCl_{4} (l) + 2 SCl_{2} (s)

You may start with any of the reactants and calculate how much of the other reactant you need to react with the given amount.

Let's start with CS_{2}: We have 27.3g of CS_{2}

we can calculate how many grams of Cl_{2 }we need to react with that 27.3g of CS_{2 }as below in one step:

27.3g of CS_{2} (1 mol CS_{2}/76.14g CS_{2}) x (4 mol Cl_{2}/1molCS_{2}) x (70.9g Cl_{2}/1mol Cl_{2}) = 102g Cl_{2}

Therefore we need only 102g of Chlorine gas, as we have more than enough (129g), Chlorine gas will be the excess reagent and **CS**_{2 }**will be the limiting reagent.**

Therefore we can calculate how much CCl_{4 }can be formed ONLY starting with our limiting reagent, CS_{2}: as below;

27.3g of CS_{2} (1 mol CS_{2}/76.14g CS_{2}) x (1mol CCl_{4}/1molCS_{2}) x (153.8g CCl_{4}/1mol CCl_{4}) = **55.1g CCl**_{4}

Note that you can always only use the limiting reagent for the calculations of product amounts, as it is the only reagent that'll get 100% consumed/reacted.