Question

A 1.268 g sample of a metal carbonate, MCO3, was treated 100.00 mL of 0.1083 M H2SO4, yielding CO2 gas and an aqueous solution of the metal sulfate. The solution was boiled to remove all of the dissolved CO2 and then was titrated with 0.1241 M NaOH. A 71.02 mL volume of the NaOH solution was required to neutralize the excess H2SO4.

a) Write the balanced chemical equation for this reaction.

b) What is the identity of the metal?

c) How many grams of CO2 gas were produced?

Answer 1

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)

b) identity of the metal = Ba

c) moles of CO2 = 0.0064232

molar mass = 44

mass of CO2 produced = 0.0064232x 44 = 0.283 g