Chemistry problem

Hardness in water is caused by dissolve calcium compounds. When heated some of these break down and deposits calcium carbonate as follows:

Ca(HCO3)2 → CaCO3 + H2O +CO2

This builds up as 'fur' on the inside of boilers. It can be removed by reaction with hydrochloric acid.

What mass of calcium carbonate would be produced from 10000 dm3 of water containing 0.356 g of calcium hydrogen carbonate per dm3 of water and what volume of 10M HCl would be needed to remove the solid calcium carbonate from the inside of the boiler?

This seems impossible. I need a little help.

Reply 1

Christophicus

OP

18

Please, i'm desperate.

Reply 2

Cirsium

19

Urgh i'm tired so this might take a while but it's really not impossible! ur probs just not getting replies cos ur on general chat not academic

Ca(HCO3)2 → CaCO3 + H2O +CO2

Ok 1000dm3 of water with 0.356g in each dm3 would have a total of 356g of Ca(HCO3)2 in total. The Mr of calcium HC is 40+1+1+12+12+48+48= 162

356 / 162 = 2.1975 mols of stuff

1 mole ---> 1 mole therefore need 2 work out the mass of 2.1975 mols of calcium carbonate

Mr of CaCO3 = 40 + 12 + 48 = 100

100 x 2.1975 = 219.75g of calcium carbonate

(plz plz plz check the Molecular masses are right because i'm tired!!!)

what volume of 10M HCl would be needed to remove the solid calcium carbonate from the inside of the boiler?

Same kinda principle. 10M HCl has 10 moles of HCl per dm3
You need to work out how many moles you will have
Then divide this by 10 and it'll tell you the volume you need
Sorry too tired to do the rest hope that this helps!!

bekaboo xx

Reply 3

Christophicus

OP

18

Bekaboo

Urgh i'm tired so this might take a while but it's really not impossible! ur probs just not getting replies cos ur on general chat not academic

Ca(HCO3)2 → CaCO3 + H2O +CO2

Ok 1000dm3 of water with 0.356g in each dm3 would have a total of 356g of Ca(HCO3)2 in total. The Mr of calcium HC is 40+1+1+12+12+48+48= 162

356 / 162 = 2.1975 mols of stuff

1 mole ---> 1 mole therefore need 2 work out the mass of 2.1975 mols of calcium carbonate

Mr of CaCO3 = 40 + 12 + 48 = 100

100 x 2.1975 = 219.75g of calcium carbonate

(plz plz plz check the Molecular masses are right because i'm tired!!!)

what volume of 10M HCl would be needed to remove the solid calcium carbonate from the inside of the boiler?

Same kinda principle. 10M HCl has 10 moles of HCl per dm3
You need to work out how many moles you will have
Then divide this by 10 and it'll tell you the volume you need
Sorry too tired to do the rest hope that this helps!!

bekaboo xx

Thanks a lot

Chemistry problem

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