Question

Calcium carbonate reacts with aqueous HCI to give CaCl2 and CO₂ according to the reaction, CaCO3 (s) + 2 HCI (aq) →→ CaCl₂ (aq) + CO₂(g) + H₂O(l) What mass of CaCO3 is required to react completely with 50 mL of 0.5 M HCI? ​

Answer 1

It can be solved with two steps :

1) first we will calculate the mass of HCl in 25ml of 0.75M HCl.

2) Now, calculate the mass of CaCO3 by using all information available from the balanced chemical equation.

 

Step 1 : calculation of mass of HCl in 25 ml of 0.75M HCl .

we know,

Molarity = mass of solute/volume of solution in L

0.75 = mass of HCL/25ml x 1000

mass of HCl = 0.6844 g

 

 

Step 2: calculation of the mass of CaCO3,

CaCO3 + 2HCl ------> CaCl2 + CO2 + H2O

here we see that,

2 moles of HCl reacts with 1 mole of CaCO3.

so, 2 × 36.5g of HCl reacts with 100g of CaCO3.

so, 73g of HCl reacts with 100g of CaCO3.

so, 1g of HCl reacts with 100/73 g of CaCO3.

so, 0.6844 g of HCl reacts with 100 × 0.6844/73g of CaCO3 = 68.44/73 g = 0.9375g

 

0.9375g of CaCO3 is required to react completely with 25ml of 0.75M HCl.