CaCO3 + 2HCL = CaCl2 + H2O + CO2

How many milliliters of 0.200 M HCL can react with 8.25g of CaCO3?

825 mL

Explanation:

We are given;

Molarity of HCl is 0.200 M Mass of CaCO₃ is 8.25 g The equation of the reaction as;

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

We are required to calculate the volume of the Acid that reacted.

Step 1: Determine the moles of CaCO₃ that reacted;

We know that;

Moles = Mass : molar mass

Molar mass CaCO₃ is 100.1 g/mol

Therefore;

Moles of CaCO₃ = 8.25 g : 100.1 g/mol

= 0.0824 moles

Step 2: Determine the moles of HCl that reacted;

From the reaction, 1 mole of CaCO₃ reacts with 2 moles of HCl

Therefore;

Moles of HCl = Moles of CaCO₃ * 2

= 0.0824 moles * 2

= 0.1648 moles

= 0.165 moles

Step 3: Determine the volume of HCl

We know that, Molarity = Moles : Volume

Rearranging the formula;

Volume = Moles : Molarity

Therefore;

Volume of HCl = 0.165 moles : 0.200M

= 0.825 L

But, 1 L = 1000 mL

= 825 mL

Therefore, the volume of HCl that reacted with CaCO₃ 825 mL