Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2

Que. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l). What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Ans. 0.75 M means 0.75 mol of HCl is present in 1 L of solution
1 Mol of HCl weighs = 1 + 35.5 = 36.5 g
0.75 mol of HCl weighs = 0.75 × 36.5 = 27.37 g
⸫ 27.37 g of HCl is present in 1000 ml (1 L) of solution
1000 ml of solution contains HCl = 27.37 g
1 ml of solution contain HCl = 27.37/1000 = .02737 g
25 ml of solution contain HCl = 0.2737 × 25 = 0.684 g
We have the chemical equation
CaCO₃ + 2 HCl à CaCl₂ + CO₂ + H₂O
2 mol of HCl (i.e 2 × 35.5 = 71 g) react with 1 mol of CaCO₃ (i.e 100 g)
⸫ 71 g of HCl reacts with CaCO₃ = 100 g
1 g of HCl reacts with CaCO₃ = 100/71 = 1.408 g
⸫ 0.684 g of HCl reacts with CaCO₃ = 1.408 × 0.684 = 0.963 g