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4 years ago

Volume of carbon dioxide gas evolved at STP by heating 7.3 gram of Mg(Hco3)2 will be

Chemistry

2 answers:

Y_Kistochka [10]4 years ago

8 0

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

krok68 [10]4 years ago

4 0

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

<h2>Further Explanation: </h2><h3>Molar gas volume </h3>

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).

<h3>Decomposition reactions </h3>

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;

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

<h3>Step 1: Moles of Mg(HCO₃)₂ in 7.3 g</h3>

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

<h3>Step 2: Moles of Carbon dioxide </h3>

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

<h3>Step 3: Volume of Carbon dioxide</h3>

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: brainly.com/question/5592681
Moles calculation: brainly.com/question/5592681
Molar mass: brainly.com/question/5592681

Level: High school

Subject: Chemistry

Topic: Moles

Sub-topic: Molar gas volume

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Volume of carbon dioxide gas evolved at STP by heating 7.3 gram of Mg(Hco3)2 will be​

Volume of carbon dioxide gas evolved at STP by heating 7.3 gram of Mg(Hco3)2 will be