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

Answer:

a) MCO3 + H2SO4 → MSO4 + CO2 + H2O

2NaOH + H2SO4 → Na2SO4 + 2 H2O

b) The metal is barium

c) There is 0.2827 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of metal carbonate = 1.268 grams

Volume of 0.1083 M H2SO4 solution = 100.00 mL

The metal sulfate 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

Step 2: The balanced equation:

MCO3 + H2SO4 → MSO4 + CO2 + H2O

2NaOH + H2SO4 → Na2SO4 + 2 H2O

Step 3: Calculate moles of H2SO4

Moles H2SO4 = molarity * volume

Moles H2SO4 = 0.1083 M * 0.100 L

Moles H2SO4 = 0.01083 moles

Step 4: Calculate moles of NaOH

Moles NaOH = 0.1241 M * 0.07102 L

Moles NaOH = 0.008814 moles

Step 5: Calculate limiting reactant

For 1 mol of H2SO4, we need 2 moles of NaOH to produce 1 mol of Na2SO4 and 2 moles of H2O

NaOH is the limiting reactant. It will completely be consumed ( 0.008814 moles).

H2SO4 is in excess. There will be consumed 0.008814/2 = 0.004407 moles of H2SO4. There will remain 0.01083 - 0.004407 = 0.006423 moles of H2SO4

Step 6: Calculate moles of MCO3

There will react 0.006423 moles of H2SO4 with MCO3

MCO3 + H2SO4 → H2O + CO2 + MSO4

For 1 mole H2SO4, we need 1 mole of MCO3

For 0.006423 moles of H2SO4, we need 0.006423 moles of MCO3, there will also be produced 0.006423 moles of CO2

Step 7: Calculate molar mass of MCO3

Molar mass MCO3 = mass MCO3 / moles MCO3

Molar mass MCO3 = 1.268 grams / 0.006423 moles

Molar mass MCO3 = 197.4 g/mol

Step 8: Calculate molar mass of the metal

Molar mass of C = 12 g/mol

Molar mass of O = 16 g/mol

Molar mass of metal = 197.4 - (12 + 3*16) = 137.4 g/mol

 

The metal with molar mass of 137.4 g/mol is barium

The metal carbonate is BaCO3

Step 9: Calculate mass of CO2 produced

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.006423 * 44.01 g/mol

Mass CO2 = 0.2827 grams