10,6 г                                               V =?

  Na2CO3    +    2HCl =   2NaCl     + CO2    + H2O

n = 1 моль                                        n = 1 моль

М = 106 г/моль                                Vm = 22,4 л/моль

m = 106 г                                         V = 22,4 л

10,6 г Na2CO3    -    X л СО2

106 г Na2CO3    -    22,4 л СО2

V(CO2) = 10,6 * 22,4 / 106 = 2,24 л