# Chemistry equation question

Could anyone help me with this chemistry equation??

Calculate the volume of chlorine needed to react with 14 g of iron .The equation for the reaction is
2Fe+3Cl2–->2FeCl3
Calculate the volume of chlorine needed to react with 14 g of iron
Fe =56
(edited 2 months ago)
Original post by Andrari20
Calculate the volume of chlorine needed to react with 14 g of iron .The equation for the reaction is
2Fe+3Cl2–->2FeCl3
Calculate the volume of chlorine needed to react with 14 g of iron
Fe =56

Original post by charco

I’m so sorry I meant to say I need help with understanding what to do
Original post by Andrari20
Could anyone help me with this chemistry equation??

Calculate the volume of chlorine needed to react with 14 g of iron .The equation for the reaction is
2Fe+3Cl2–->2FeCl3
Calculate the volume of chlorine needed to react with 14 g of iron
Fe =56

Do you know what the balanced equation means?

2Fe+3Cl2 ==> 2FeCl3

This says that 2 mol of iron will react with 3 mol of chlorine to produce 2 mol of iron(III) chloride.
This ratio is ALWAYS respected.

Do you know how to calculate mol from mass? You need that skill for this question.

Mol = mass/relative mass

Now try the question yourself.
I understand the equation moles =mass/Rfm
but I’m just not sure if I’ve got my number correct at all but I used
14 g /112 =0.125 mol and then I’m completely lost in what to do next even if that’s right which I don’t think it is ???
(edited 1 month ago)
Original post by Andrari20
I understand the equation moles =mass/Rfm
but I’m just not sure if I’ve got my number correct at all but I used
14 g /112 =0.125 mol and then I’m completely lost in what to do next even if that’s right which I don’t think it is ???

You should not use the equation coefficients when determining moles.

Moles of iron = 14/56 = 0.25 mol
From the equation stoichiometry (the coefficients) you can see that 2 mol of iron makes 2 mol of iron(III) chloride.

So, however many mol of iron you have initially, you must end up with the same mol of iron(III) chloride.
You start with 0.25 mol iron so you end up with 0.25 mol of iron(III) chloride.

The relative formula mass of iron(III) chloride = 56 x 3(35.5) = 162.5
So you end up with 0.25 x 162.5 g of iron(III) chloride = 40.625.

To know the mass of chlorine used you apply exactly the same process.
Have a go!
If you're still struggling, here's a video I made many moons ago ...
this page has similar type mole calculations, the examples maybe different but all these mole calculations are done the same way https://www.science-revision.co.uk/moles%20and%20equations.html
Original post by charco
You should not use the equation coefficients when determining moles.

Moles of iron = 14/56 = 0.25 mol
From the equation stoichiometry (the coefficients) you can see that 2 mol of iron makes 2 mol of iron(III) chloride.

So, however many mol of iron you have initially, you must end up with the same mol of iron(III) chloride.
You start with 0.25 mol iron so you end up with 0.25 mol of iron(III) chloride.

The relative formula mass of iron(III) chloride = 56 x 3(35.5) = 162.5
So you end up with 0.25 x 162.5 g of iron(III) chloride = 40.625.

To know the mass of chlorine used you apply exactly the same process.
Have a go!

The answer I’ve got is 9dm3 does that sound correct
Original post by Andrari20
The answer I’ve got is 9dm3 does that sound correct

Yes