A 1.268 g sample of a metal carbonate, MCO33, was treated 100.00 mL of 0.1083 M...

A 1.268 g sample of a metal carbonate, MCO33, was treated 100.00 mL of 0.1083 M H22SO44, yielding CO22 gas and an aqueous solution of the metal sulfate. The solution was boiled to remove all of the dissolved CO22 and then was titrated with 0.1241 M NaOH. A 71.02 mL volume of the NaOH solution was required to neutralize the excess H22SO44.

Start by writing a balanced chemical equation for this reaction.

What's the identity of the metal?

Give the elemental symbol

How many grams of CO2 gas were produced?

Solutions

Expert Solution

Identity of the metal : Ba

Grams of CO2 produced : 0.2827g

Explanation

2NaOH(aq) + H2SO4(aq) ---------> Na2SO4(aq) + 2H2O

Stoichiometrically, 2moles of NaOH react with 1mole of H2SO4

moles of NaOH consumed = (0.1241mol/1000ml) × 71.02ml = 0.0088136mol

moles of excess H2SO4 = 0.0088135mol/2 = 0.0044068mol

Total moles of H2SO4 = ( 0.1083mol/1000ml)×100ml = 0.01083mol

moles of H2SO4 reacted with MCO3 = 0.01083mol - 0.0044068mol = 0.0064232mol

MCO3(s) + H2SO4(aq) ---------> MSO4(aq) + CO2(g) + H2O(l)

moles of CO2 produced = 0.0064232mol

moles of H2O produced = 0.0064232mol

moles of M²⁺ = 0.00644232mol

mass of CO2 produced = 0.0064232mol × 44.01g/mol = 0.2827g

mass of H2O produced = 0.0064232mol × 18.016g/mol = 0.1157g

mass of M²⁺ = 1.268g - ( 0.2827g + 0.1157g) = 0.8696g

molar mass the metal = 0.8696g/0.0064232mol

molar mass of the metal = 135.4g

As per the molar mass of the metal , the metal is Barium and the sympol is Ba

queen_honey_blossom answered 8 months ago

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