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Help with Chemistry!

The overall equation for the Solvay Process is
2NaCl + CaCO3 → Na2CO3+ CaCl2
Calculate the maximum mass of sodium carbonate that could be formed by
reacting 40 kg of calcium carbonate with an excess of sodium chloride solution.
(Relative formula masses: CaCO3 = 100; Na2CO3 = 106)

help I'm not sure how to work this outtt, my exam is today!!!

Reply 1

Wimsett

Use 'What is the MASS of MR MOLE RATIO?'

Write down the:
Mass
Mr
Moles
Ratio

Then use equations to find what's missing

Reply 2

CorpusLuteum

Original post by Wimsett

Use 'What is the MASS of MR MOLE RATIO?'

Write down the:
Mass
Mr
Moles
Ratio

Then use equations to find what's missing

The thing is, moles isn't part of C2.
And it's confusing the **** out of me. D:
Could you lay it out for me please?

Reply 3

Wimsett

Original post by CorpusLuteum

The thing is, moles isn't part of C2.
And it's confusing the **** out of me. D:
Could you lay it out for me please?

I'm pretty sure moles is a part of c2, but I do triple so maybe I just get them mixed up. Do you do AQA?

Ok so first write down the mass and mr (which they already give) for each thing, then use what you have to find the moles for what you can. Look at the ratio in the equation as you would do for titration and use this to work out what's missing.

Just draw a table and fill very thing you can in, and work your way through it using moles = mass/mr

Reply 4

z1820

Original post by CorpusLuteum

Mr: (relative formula mass)
100 -------> 106

What do you do to a 100 to get to 106? X 1.06

Whatever you do to the Mr, do to the mass so:

Mass:
40kg ------> ?

40 X 1.06 = 42.4 kg

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Reply 5

CorpusLuteum

Original post by z1820

But what calculation do you do to find out how you got from 100 to 106?
D:

Original post by Wimsett

I do edexcel. I still Can't get that method down right.

Reply 6

z1820

Original post by CorpusLuteum

But what calculation do you do to find out how you got from 100 to 106?
D:

Just 106 divided by 100
Which is 1.06 :smile:

I do edexcel. I still Can't get that method down right.

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Reply 7

CorpusLuteum

Original post by z1820

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Oh thank you so much!!!
:biggrin:

Reply 8

z1820

Original post by CorpusLuteum

Oh thank you so much!!!
:biggrin:

No problem - good luck for the exam!
:smile:

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Reply 9

DrPeter

Original post by CorpusLuteum

I know you have received the correct answer, and if you are doing AQA the method is sufficient. Here is the full working, just in case you, or anyone else is doing the Edexcel GCSE/ IGCSE (where moles should be understood) and also if the mole ratio of reactant to product is different in the question.

Moles CaCO3 = mass (g)/MR = 40000/100 =400 moles

Mole Ratio CaCO3:Na2CO3 = 1:1

Therefore, 400 moles CaCO3 would produce 400 moles Na2CO3

Therefore mass Na2CO3 = moles x MR = 400 x 106 = 42,400g =42.4 Kg

Hope this helps somebody !!

Reply 10

CorpusLuteum

Original post by DrPeter

Thank you, this will be a good refrence for Triple science! :biggrin:

Reply 11

Aya Arwadi

Original post by CorpusLuteum

2NaCl + CaCO3 → Na2CO3+ CaCl2
100 g 106 g
40,000g ?
100 and 106 and 40 kg from given.
using cross multiplication,( 40000 x 106 )/ 100 is equal to 42400 g of sodium bicarbonate, which is 42.4 kg.