Unit 6 Test Review

Question

Elemental iron can be formed from iron (III) chloride and elemental lithium. How many grams of iron can be made from 23.1g of lithium and 45g of iron (III) chloride?

Li + FeCl₃ → LiCl + Fe (unbalanced)



23.1

 

45

Which reactant is the limiting reactant? Enter a 1 for lithium or a 2 for iron (III) chloride. ____

How many gram of iron are actually produced? ___g

Answer 1

Hi, Sam,

We need to start with a balanced equation. It is hard to tell whether this is where you were having difficulty - please indicate which part of the problem you'd like help on. Post a balanced equation, if you got that far.

I can explain more about balancing, but will simply summarize it here, due to lack of time, space, and appreciation of which part of the question is unclear.

3Li + FeCl₃ → 3LiCl + Fe

Next, we calculate the moles we have of each reactant:

Li: 14.7g/6.94g/mole = 2.12 moles.

FeCl₃ : 45g/162.2g/mole = 0.277 moles FeCl₃

We see from the equation that one mole of FeCl₃ requires 3 moles of lithium. Ugh. Lithium is a spry, lively element that is a joy to have around. Iron is a thug who runs off with any oxygens it can find. Too bad lithium is being used this way.

The upside is that there isn't much FeCl₃ available. The 0.2774 moles of it will consume 3 times that many moles of Lithium, or 0.832 moles Li. Since we have 2.12 moles of Li, we will consume all the FeCl₃ and leave behind 1.29 moles of lithium. So the iron chloride is the limiting reagent.

To find grams iron produced, we see that there is 1 mole Fe for every 1 mole FeCl₃ consumed. We used all the 0.277 moles of FeCl₃, so we'll produce 0.277 moles of Fe

(0.277 moles Fe)(55.85g/mole) = 15.49 grams Fe

Bob