Respuesta :
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
The volume of carbon dioxide gas evolved at STP by heating 7.3 gram of Mg(HCO3)2 will be 1.12 L or 1120 ml
Further Explanation:
Molar gas volume
- Molar gas volume states that one mole of a gas occupies a volume of 24 liters when the gas is at room temperature and pressure (rtp), while one mole of a gas will occupy a volume of 22.4 liters at standard temperature and pressure (STP).
Decomposition reactions
- These are reactions that involve break down of a compound to small molecules or atoms. Decomposition may be carried using heat (thermal decomposition) or using a catalyst (catalytic decomposition).
- An example;
The reaction is a decomposition reaction;
Mg(HCO₃)₂ + Heat → MgCO₃ + H₂O + CO₂
From the reaction
1 mole of Mg(HCO₃)₂ produces 1 mole of CO₂ after decomposition.
Therefore, to determine the volume of CO₂ produced by 7.3 grams of Mg(HCO₃)₂
Step 1: Moles of Mg(HCO₃)₂ in 7.3 g
Number of moles = Mass/relative formula mass
RFM of Mg(HCO₃)₂ =146.33868 g/mol
Therefor;
Number of moles = 7.3 g/146.33868 g/mol
= 0.04988
= 0.05 moles
Step 2: Moles of Carbon dioxide
From the equation;
1 mole of Mg(HCO₃)₂ produces 1 mole of CO₂ after decomposition.
Therefore;
Moles of carbon dioxide produced is 0.05 moles
Step 3: Volume of Carbon dioxide
From molar gas volume;
1 mole of CO₂ at STP occupies 22.4 L
Thus;
0.05 moles will occupy;
= 0.05 × 22.4 L
= 1.12 L or 1120 mL
Therefore; volume of carbon dioxide gas evolved at STP by heating 7.3 gram of Mg(Hco3)2 will be 1.12 L or 1120 ml
Keywords; Decomposition reactions, molar gas volume, relative formula mass, moles
Learn more about:
- Relative Formula mass: https://brainly.com/question/5592681
- Moles calculation: https://brainly.com/question/5592681
- Molar mass: https://brainly.com/question/5592681
Level: High school
Subject: Chemistry
Topic: Moles
Sub-topic: Molar gas volume
