chem homework help urgent

Question

For the following reaction, 27.3 grams of carbon disulfide are allowed to react with 129 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?  grams

What is the FORMULA for the limiting reagent?

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

Answer 1

First you need the balanced Equation as below:

CS₂ (s) + 4 Cl₂ (g) → CCl₄ (l) + 2 SCl₂ (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 CS₂: We have 27.3g of CS₂

we can calculate how many grams of Cl₂ we need to react with that 27.3g of CS₂ as below in one step:

27.3g of CS₂ (1 mol CS₂/76.14g CS₂) x (4 mol Cl₂/1molCS₂) x (70.9g Cl₂/1mol Cl₂) = 102g Cl₂

Therefore we need only 102g of Chlorine gas, as we have more than enough (129g), Chlorine gas will be the excess reagent and CS₂ will be the limiting reagent.

Therefore we can calculate how much CCl₄ can be formed ONLY starting with our limiting reagent, CS₂: as below;

27.3g of CS₂ (1 mol CS₂/76.14g CS₂) x (1mol CCl₄/1molCS₂) x (153.8g CCl₄/1mol CCl₄) = 55.1g CCl₄

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.