In a certain experiment, 20.00 mL of 0.0800 M HCL were added to a 0.1022-g sample of MCO3. The excess HCL required 5.64 mL of 0.1000 M NaOH for neutralization. Calculate the molar mass of the carbonate and identify M?

HCl:                  20 ml x 0,0800 M = 1,6 meq. (milliequivalents)

NaOH:                5.64 ml x 0,1000 M=0,564 meq. -

Sample                                                1,036 meq

1 mmol carbonate equals 2 meq.

1 mmol carbonate = 1 mmol MCO3.

mmol carbonate  = 1,036/2 = 0,518.

mmol carbonate = weight (mg)/Molar mass

        0,518                    = 102,2/M

M = 197,29

Molar mass carbonate = 3 x O + 1 x C = 3 x 16 = 1 x 12 = 60

atomic weight metal = 197,29 - 60 = 137,29 = Barium