Question

Question 1.35 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?

Class XI Some Basic Concepts of Chemistry Page 25

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 balance 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
$0.75=\frac{mass\:of\:HCl}{25ml}\times\:1000$
mass of HCl = 0.6844 g


step 2 : calculation of mass of CaCO3 ,
CaCO3 + 2HCl ------> CaCl2 + CO2 + H2O
here we see that,
2 mole 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.

Answer 2

Mass of CaCO3 required is 0.94 grams

Explanation:

CaCO3 + 2HCl  —> CaCl2 + CO2 + H2O

No of moles of HCl given = MhclVhcl

= 0.75 mole $L^{-1}$ * 25 * $10^{-3}$ L

= 18.75 * $10^{-3}$ mole = 0.188 mole

2 moles of HCl requires 1 mole of CaCO3

0.0188 mole of HCl will require  = 0.0188 / 2

= 0.0094 mole of CaCO3

Molar mass of CaCO3 = 100 g/mole

Mass of CaCO3 required = 100 * 0.0094 = 0.94 g