A 1.268 g sample of a metal carbonate (MCO3) was treated with
100.00mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an
aqueous solution of the metal sulfate (MSO4). The solution was
boiled to remove all the dissolved CO2 and was then titrated with
0.1241 M NaOH. A 71.02 mL volume of NaOH was required to
neutralized the excess H2SO4.
What is the identity of the metal M?
The answer is M = Ba (Barium)
MCO3 + H2SO4 => MSO4 + CO2 + H2O
2 NaOH + H2SO4 => Na2SO4 + 2 H2O
Moles of NaOH = volume x concentration
= 71.02/1000 x 0.1241 = 0.0088136 mol
Moles of excess H2SO4 = 1/2 x moles of NaOH
= 1/2 x 0.0088136 = 0.0044068 mol
Initial moles of H2SO4 = volume x concentration
= 100/1000 x 0.1083 = 0.01083 mol
Moles of reacted H2SO4 = initial moles of H2SO4 - excess moles of H2SO4
= 0.01083 - 0.0044068 = 0.0064232 mol
Moles of MCO3 = moles of reacted H2SO4 = 0.0064232 mol
Molar mass of MCO3 = mass/moles
= 1.268/0.0064232 = 197.4 g/mol
Molar mass of M = moles mass of MCO3 - molar mass of CO3
= 197.4 - 60.0 = 137.4 g/mol
From the periodic table, the element with molar mass of 137.4 g/mol is Ba (Barium)
Thus M = Ba (Barium)
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