(CaCO₃ + Na₂CO₃) mixture = 103 g

Let moles CaCO₃ = x    

moles of Na₂CO₃ = y

  x             y             1

CaCO₃ +  Na₂CO₃ + HCl → CO₂ and other products.

CO₂ produced have C atoms from Na₂CO₃ & CaCO₃

moles of CO₂ produce = PV = nRT

1 × 24.63 = n × 0.0821 × 300

n =1

Apply POAc on C

1 × x + y × 1 = 1

x + y = 1

100 x + 106 y = 103

6x = 3 ;     x = 0.5 ;    y = 0.5

weight of CaCO₃ = 50 g

weight of Na₂CO₃ = 53 

% CaCO₃ = $\frac{50}{103}\times 100$ = 48.54 %