[T]

Hello, I tried solving this problem from the problem of the week.

Mixture of CaO and CaCO3 was reacted stoichiometrically with a solution of hydrochloric acid, producing 8.04 L of CO2 (measured at 290 K and 120 kPa). Mass of the dry CaCl2 obtained was 50.61% higher than the mass of the original mixture. How much water was present in the 1M HCl solution used for the reaction, if its density was 1.02 g/mL?

But I cannot seem to find the right chemical equation as it doesn't give me the right answer.

My equation was:

CaO + CaCO3 + 4HCl  2CaCl2 + CO2 + 2H20

I tried to solve the question by using PV= nRT and finding the mol of CO2. Then I use stoichiometry to determine the mol of HCl. Afterwards find the amount of water by n/amount of water = molarity. And then the mass of water by mass of water/amount of water = 1.02 g/mL. But the mass is not the same as the answer.

Could someone tell me how to find the right chemical equation? I might have interpreted the question wrongly too.

Thanks

[Dan]

I tried to solve the question by using PV= nRT and finding the mol of CO2. Then I use stoichiometry to determine the mol of HCl. Afterwards find the amount of water by n/amount of water = molarity. And then the mass of water by mass of water/amount of water = 1.02 g/mL. But the mass is not the same as the answer.

Please show your working.

[Arkcon]

My equation was:

CaO + CaCO3 + 4HCl  2CaCl2 + CO2 + 2H20

I would do something differently.  Write two chemical equations, one for CaCO₃ and another for CaO.  Look at the similarities, and the differences.  They're both important.

[T]

Please show your working.

PV = nRT

120 x 8.04 = n x 8.314 x 290

n = 0.4

CaO + CaCO3 + 4HCl  2CaCl2 + CO2 + 2H20

1 CO2 : 4 HCl

n(HCl) = 1.6

1 M = n(HCl)/amount of water (L)

amount of water (L) = 1.6

density (g/mL) = mass of water (g)/amount of water (mL)

1.02 = mass of water (g)/1600

mass of water (g) = 1632g

I would do something differently.  Write two chemical equations, one for CaCO₃ and another for CaO.  Look at the similarities, and the differences.  They're both important.

I tried doing that and the 2 equations are:

CaCO3 + 2HCl  CaCl2 + H2O + CO2

CaO + 2HCl  CaCl2 + H2O

And then I add them together to give the equation I have.

CaO + CaCO3 + 4HCl  2CaCl2 + CO2 + 2H20

Thanks

[Borek]

I tried doing that and the 2 equations are:

CaCO3 + 2HCl  CaCl2 + H2O + CO2

CaO + 2HCl  CaCl2 + H2O

And then I add them together to give the equation I have.

CaO + CaCO3 + 4HCl  2CaCl2 + CO2 + 2H20

When you add equations it is equivalent of assuming molar ratio of CaO and CaCO₃ is 1:1. In general, for a mixture of unknown composition, you can't do that.

[T]

I see thanks Borek.

So using those equations I worked things out:

CaCO3 + 2HCl  CaCl2 + H2O + CO2

PV = nRT
120 x 8.04 = n x 8.314 x 290
n(CO2) = 0.4
Using stoichiometry n(HCl) used in this equation is 0.8

Then to find n(HCl) used in this equation
CaO + 2HCl  CaCl2 + H2O

I wrote mass of original mixture = 0.4 x 100.09 + w x 56.08
where w = mol of CaO
mass of CaCl2 = (w + 0.4) x 110.17

Then I can write an equation
1.5061 x (0.4 x 100.09 + w x 56.08) = (w + 0.4) x 110.17
w = 0.631323

So therefore the n(HCl) used in the second equation is 1.262646 and adding the 0.8 from the first equation gives me 2.0626 mol of HCl. This means in a 1 M there will be 2.0626 L of water. Then since density = 1.02.

mass of water/2062.6 = 1.02
mass of water = 2103.90

The answer is still wrong but I think the answer lies in Rutherford who replied to the original post. He wrote:

"and that the mass share is 3.5784% (4 digits). So:
73g:x=3.5784%:(100-3.5784)%, x=1967.02g of water."

Could someone explain this?
Thanks

[Borek]

gives me 2.0626 mol of HCl. This means in a 1 M there will be 2.0626 L of water.

No.

And check your molar mass of CaCl₂.

In the future please post link to the original thread. Sometimes possible problems with these questions were already discussed.

[T]

Thanks Borek, I will post these questions on the post in the future.

No.

I don't understand this. I changed the molar mass and with the new amount of HCl it still doesn't give me 1965 g. I might have wrote that badly, so I will reword it: Since the solution is 1 M, then if there is 2.06 mol of HCl then there will be 2.06 L of water. Could someone explain this to me?

Thanks

[Borek]

Since the solution is 1 M, then if there is 2.06 mol of HCl then there will be 2.06 L of water.

Not of the water - of the solution.

[T]

Ok I see now, thank you very much Borek and everyone who helped me.