Question

In: Chemistry

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?

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 1 year ago

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