Fe + Cl2 --> FeCl3 : Calculations & Excess Reactant

[sues]

1.0 g of Iron metal reacts with 2.0 g of chlorine gas to make iron (III) chloride, FeCl3.

(a) Write a balanced equation for this reaction.
(b) Which of the reactants is present in excess ?
(c) What mass of this reactant will be left over at the end of the reaction ? (7 marks)

Homework Equations

OK with equation : 2Fe + 3Cl2 = 2FeCl3

The Attempt at a Solution

Went for moles of Fe as 2/112 and moles of Cl2 as 2/213 then didn't know what to do with numbers

 

[tiny-tim]

Welcome to PF!

1.0 g of Iron metal reacts with 2.0 g of chlorine gas to make iron (III) chloride, FeCl3.

(a) Write a balanced equation for this reaction.
(b) Which of the reactants is present in excess ?
(c) What mass of this reactant will be left over at the end of the reaction ? (7 marks)

Homework Equations

OK with equation : 2Fe + 3Cl2 = 2FeCl3

The Attempt at a Solution

Went for moles of Fe as 2/112 and moles of Cl2 as 2/213 then didn't know what to do with numbers

Hi sues! Welcome to PF!

Hint: how many moles in 1g of iron and in 1g of chlorine?

 

[sues]

Thanks Tiny Tim.
So if I did the following am I on the right track?
Moles of Fe = 1/112 and moles of Cl2 = 2/213 (taking into account stoiciometry).
Therefore Cl2 in excess.

Mass of Cl2 left after reaction:

9.39x10-3 - 8.92x10-3 = 4.7x10-4

Mass = 4.7x10-4 x 71 = 0.0334g


Any good?

 

[GCT]

Thanks Tiny Tim.
So if I did the following am I on the right track?
Moles of Fe = 1/112 and moles of Cl2 = 2/213 (taking into account stoiciometry).
Therefore Cl2 in excess.

Mass of Cl2 left after reaction:

9.39x10-3 - 8.92x10-3 = 4.7x10-4

Mass = 4.7x10-4 x 71 = 0.0334g


Any good?

Where are you getting these numbers?

- Convert one of the reagents into moles

- Next find out the mole equivalence of the other reagent e.g. if you used Chlorine gas then note that there is two Iron III for every three Chlorine gases , use this ratio to find the moles of Chlorine then convert this value into grams

- Compare this value with the one given to you to understand it is going to suffice then you know the limiting reagent.

Show your work then we can move on to the second part.

 

[sues]

Ok here goes!
moles of Cl2= 2/71 = 0.028
moles of Fe = 2/3 x 0.028 = 0.0187
mass of Fe = 0.0187 x 57 = 1.064g

or other way

moles of Fe = 0.018
moles of l2 = 0.027
mass of Cl2 = 1.917g therefore Cl2 in excess.

mass of reactant in excess left over = 2.0 - 1.917 = 0.083g

Thanks

 

[Borek]

OK, although your numbers are slightly off, you have either used some strange molar masses or rounded intermediate results.

Red means in excess.

 

[GCT]

Ok here goes!
moles of Cl2= 2/71 = 0.028
moles of Fe = 2/3 x 0.028 = 0.0187
mass of Fe = 0.0187 x 57 = 1.064g

or other way

moles of Fe = 0.018
moles of l2 = 0.027
mass of Cl2 = 1.917g therefore Cl2 in excess.

mass of reactant in excess left over = 2.0 - 1.917 = 0.083g

Thanks

Great! Just be certain to round off those numbers correctly. BTW I haven't checked on whether the numbers for the second portion are correct since you did not display your work here.