Limestone rock contains solid calcium carbonate that can react with

acid to give products. In one experiment, 1.452 g of limestone rock is

pulverised and then treated with 25.00 mL of 1.035 M HCl solution.

After the reaction is completed, the resulting mixture is still acidic. The

excess acid then requires 15.25 mL of 0.1010 M NaOH for

neutralisation. Assuming that calcium carbonate is the only substance

in limestone rock that reacts with HCl solution:

i. Write the balanced chemical equation and net ionic equation for

the reaction that occurs between calcium carbonate and

hydrochloric acid.

[2 marks]

10

FAC 0015

ii. Calculate the number of moles of hydrochloric acid that reacts

with calcium carbonate.

[3 marks]

iii. Calculate the percent by mass of calcium carbonate in the rock.

1 answer

Limestone rock contains solid calcium carbonate that can react with

acid to give products. In one experiment, 1.452 g of limestone rock is

pulverised and then treated with 25.00 mL of 1.035 M HCl solution.

After the reaction is completed, the resulting mixture is still acidic. The

excess acid then requires 15.25 mL of 0.1010 M NaOH for

neutralisation. Assuming that calcium carbonate is the only substance

in limestone rock that reacts with HCl solution:

i. Write the balanced chemical equation and net ionic equation for

the reaction that occurs between calcium carbonate and

hydrochloric acid.

[2 marks]

10I don't understand this 10

FAC 0015 I don't understand this number either
CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O
CO3^2- + 2H^+ ==> CO2 + H2O


ii. Calculate the number of moles of hydrochloric acid that reacts

with calcium carbonate.

[3 marks]
mols HCl added initially - mols HCl added in excess = mols HCl reacted with sample.
Initial mols HCl added = M HCl x L HCl = ?.
mols HCl in excess = mols NaOH x L NaOH = ?
Subtract to find mols HCl used in the reaction.


iii. Calculate the percent by mass of calcium carbonate in the rock.

Convert mols HCl used in the reaction to mols CaCO3 reacted. That is mols HCl x (1 mol CaCO3/2 mols HCl) = ?
Then grams CaCO3 = mols CaCO3 x molar mass CaCO3.
Finally, %CaCO3 = (g CaCO3/mass initial sample)*100 = ?