Yr 11: Can anybody help me with this question? (1 Viewer)

[chobui11]

Egg laying hens are fed a calcium supplement in the form of shell grit, consisting of crushed sea shells and sand. To investigate shell grit for calcium carbonate content, a 10 g sample is mixed with excess hydrochloric acid. The mass of the reaction mixture is monitored using an electronic balance. The mixture decreases by 2.75 g as the reaction proceeds.

a. Account for the change in mass of the reaction mixture.

b. Write a balanced equation for the mixture.

c. Calculate the mass AND percentage by mass of the calcium carbonate in the sample

d. Describe a suitable procedure to separate a dried sample of the unreacted residue, for further analysis.

Thanks

 

[Axio]

a. As the mixture has 'lost' some of its mass, a gas must have been produced through a chemical reaction.

b. I think this might mean balanced equation for the reaction NOT mixture... so assuming that:

CaCO3 + 2HCl -> CaCl2 + H2CO3

And I had to look this up but apparently H2CO3 is unstable in water at high concentrations so it becomes:
H2CO3 -> H2O + CO2 or CaCO3 + 2HCl -> CaCl2 + H2O + CO2

c. So, 2.75g of mass is lost = mass of CO2 in reaction, so the number of moles in the reaction is: 2.75g/(12+16 x 2 g/mol) = 0.0625mol

CaCO3 : CO2 is:
1 mole : 1mole

Therefore 0.0625mol of CaCO3 is in the reaction.

So the mass of CaCO3 = 0.0625 x (40.1+12+16 x 3) = 6.25625g = 6.26g

And the percentage mass = ((10-6.25625)*100)/10 = 37.4%

d. Now you have to come up with a procedure to separate Calcium chloride and Water.

(I am not 100% sure about reactions in b. but I think my method is correct)

 

[zhertec]

A. The gas released is due to the reaction between the HCl and CaCO3, resulting in a neutralisation reaction between the acid and carbonate, the gas released being CO2.

B. So when an acid reacts with a carbonate the reaction is: 2HCl (aq) + CaCO3 (s) --> CaCl2 (aq) + CO2 (g) + H2O (l)

C. 2.75g mass is lost due to CO2 being released. so convert that into moles of CO2 released: 2.75g/(12.01 + (16x2)) = 0.0625 moles rounded.
Since that is a 1:1 mole ratio with CaCO3: M/MM(CaCO3) = 0.0625
Therefore M=0.0625 x (40.08+12.01+(16x3)) = 6.256g of CaCO3 present.
6.256g/10g = around 62.56% of it was CaCO3

D. Now you have a solution of CaCl2 (aq), Excess HCl, Water and possibly dissolved CO2 (minuscule amount) with some physical residue that has not been reacted (so still solid I believe) so the best way is to filter the mixture to obtain the residue, (as everything else is either in aqueous or liquid which will pass right through the filter paper) and dry the residue.

P.S. Correct me if I got something wrong in that answer :3

 

[someth1ng]

A. The gas released is due to the reaction between the HCl and CaCO3, resulting in a neutralisation reaction between the acid and carbonate, the gas released being CO2.

Consider referring to why the mass loss. Try to use keywords in the question to make your answer seem more relevant.

 

[zhertec]

The changes in mass was due to the reaction between an Acid (HCl) and a carbonate (CaCO3). When an acid and carbonate reacts, it produces Carbon dioxide gas which escapes from the solution resulting in the loss of products produced, hence mass.

Is that more or less correct....? lol sorry if I got it wrong again.