For the following reaction, 184 grams of iron are allowed to react with 215 grams of chlorine gas.

Question

For the following reaction, 184 grams of iron are allowed to react with 215 grams of chlorine gas.

2Fe (s) + 3Cl₂ (g) 2FeCl₃ (s)

What is the FORMULA for the limiting reagent?

What is the maximum amount of iron(III) chloride that can be formed?  in grams

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

Answer 1

Hello, Dayer,

First find the moles of both reactants. Divide the masses by the respective molar masses to obtain moles. The balanced equation tells us we need 2 moles of iron for every 3 moles of Cl₂: (2moles Fe)/(3 moles Cl₂). We have more than enough Fe to conume all the Cl₂. The Cl₂ is the limiting reagent. We consume only 2.02 moles of Fe.

The balanced equation says we get one mole of FeCl₃ for every mole of Fe consumed. So we obtain 2.02 moles of FeCl₃. Multiply the molar mass of FeCl₃ by the moles to obtain grams FeCl₃.