Volume of carbon dioxide gas evolved at stp by heating 7.3 gram of mg(hco3)2 will be

1.12dm³

Explanation:

The reaction is a decompositon reaction. It is expressed below:

Mg(HCO₃)₂ + Heat → MgCO₃ + H₂O + CO₂

This is the balanced reaction equation.

We have 1 mole of Mg, 2 mole of H, 2 mole of C, 6 mole of O on both sides of the reaction.

To calculate the volume of CO₂ gas evolved at S.T.P:

We work from the known to the unknown.

The known is Mg(HCO₃)₂ and the unknown is CO₂

1. Find the number of moles of Mg(HCO₃)₂ from the given mass

2. Since we know that chemical reactions obey the law of conservation of mass. The balanced chemical reaction shows that, 1 mole of Mg(HCO₃)₂ would give 1 mole of CO₂

3. We use information from (2) to find the number of moles of CO₂. With the number of moles of CO₂ found, we use the formula below to find the volume evolved at S.T.P

Volume evolved =

number of moles of CO₂ x 22.4dm³mol⁻¹

Solution:

Given/known parameters:

Mass of Mg(HCO₃)₂ = 7.3g

Step 1:

Number of moles

of Mg(HCO₃)₂ = mass/molar mass

Molar mass of Mg(HCO₃)₂ :

Atomic mass of Mg = 24gmol⁻¹

H = 1gmol⁻¹

C = 12gmol⁻¹

O = 16gmol⁻¹

Molar Mass Mg(HCO₃)₂

= [24 + 2{1 + 12 + (16 x3)} ]

= 24 + 2(1 +12 + 48)

= 24 + 122

= 146gmol⁻¹

Number of moles of

Mg(HCO₃)₂ = 7.3g/126gmol⁻¹ = 0.05mol

Step 2:

We know from the balanced equation that:

1mole of Mg(HCO₃)₂ = 1 mole of CO₂

So: 0.05mol of Mg(HCO₃)₂ = ?

This would also produce 0.05mol of CO₂

Step 3:

Volume of CO₂ evolved =

number of moles of CO₂ x 22.4dm³mol⁻¹

= 0.05 x 22.4

= 1.12dm³

The volume of CO₂ gas evolved at S. T. P is 1.12dm³

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