Answer to Question #166213 in General Chemistry for TAeeera

Question

For the following reaction, 166 grams of iron are allowed to react with 313 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?  ..........grams

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

Please with steps, Thanksss.

Answer 1

1 We should calculate number of moles of each reactant

$$n=m/M$$

$$n (Fe)= 166/56=2.97 mol$$

$$n(Cl2)=313/71=4.4 mol$$

2 According to the chemical reaction there is the ratio

$$2.97/2=4.4/3$$

$$1.485=1.467$$

3 The limiting reagent is the one which is less. So it is Cl2

4

$$4.4/3=n (FeCl3)/2$$

$$n(FeCl3)=4.4*2/3=2.93 mol$$

$$m(FeCl3)=nM=2.93*162.5=176.125 g$$

5 Amount of the excess reagent remains after the reaction is complete

Amount of the Fe that react is

$$n=2*4.4/3=2.93 mol$$

Fe that remains

$$n=2.97-2.93=0.04 mol$$

$$m=0.04*56=2.24 g$$