Answer to Question #166213 in General Chemistry for TAeeera

Question #166213

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.


1
Expert's answer
2021-02-28T06:18:29-0500

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"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS