Question

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.

Answer 1

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<b>I don't understand this 10</b>

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

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

with calcium carbonate.

[3 marks]
<b>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. </b>

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

<b>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 = ?</b>