Answer:
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
