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

2Fe+3Cl2–->2FeCl3

Calculate the volume of chlorine needed to react with 14 g of iron

Fe =56

... and your ideas are?

Original post by charco

... and your ideas are?

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

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+3Cl

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.

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 ???

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!

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

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