This problem is confusing. Is that second titration a 25 mL aliquot of the original or has the solution been boiled so that the new volume is only 25 mL. If the latter, then
MCO3 + 2HCl ==> MCl2 + CO2 + H2O
mols HCl initially = 0.250 L x 0.203 M = 50.75/1000 = ?
mols NaOH added = mols HCl in excess = 0.0241 x 0.1M = ?
HCl used in the reaction = initial-excess
mols MCO3 used in the rxn = mols HCl/2 (from the coefficients in the balanced equation).
Then molar mass MCO3= grams/mols
Atomic mass M = molar mass MCO3-atomic mass C - 3*atomic mass O.
1.350g of insoluble carbonate, MCO3 was dissolved in 250cm^3 of a 0.203 mol/dm^3 HCL solution. The resulting solution was boiled off to remove all carbon dioxide produced. 25.0 cm^3 of the solution required 24.10cm^3 of 0.100 mol/dm^3 sodium hydroxide solution. Determine the relative atomic mass of M
1 answer
1 answer