Solve Volume of Reactant H2SO4 for MgCO3 Reaction

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In summary, A certain volume of 0.800M sulfuric acid was added to solid magnesium carbonate, and the carbon dioxide evolved was passed through concentrated sodium hydoxide solution. When the reaction had ceased, the mass of this solution was found to have increased by 15.0g.
  • #1
pivoxa15
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Homework Statement


A certain volume of a solution of 0.800M sulfuric acid was added to solid magnesium carbonate, and the carbon dioxide evolved was passed through concentrated sodium hydoxide solution. When the reaction had ceased, the mass of this solution was found to have increased by 15.0g

What volume of sulfuric acid must have been added?




Homework Equations


Basic stoichiometry


The Attempt at a Solution


H2SO4 + MgCO3 -> MgSO4 + H2O + CO2
CO2 + NaOH -> NaHCO3

Hence
H2SO4 + MgCO3 -> MgSO4 + H2O + NaHCO3

The extra weight comes from NaOH. Since the weight of the products had increased by 15g it must be due to NaOH. It comes out to .375 moles of NaOH.

Since the mole ratio for every molecule is 1, there should be .375 moles of H2SO4 added. There are 0.8 moles of H2SO4 per litre so .375/.8=.47L or 470ml of H2SO4 but the answer at the back of the book suggest something else. I don’t know where I could have made my mistake.
 
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  • #2
pivoxa15 said:
...and the carbon dioxide evolved was passed through concentrated sodium hydoxide solution. When the reaction had ceased, the mass of this solution was found to have increased by 15.0g


The extra weight comes from NaOH. Since the weight of the products had increased by 15g it must be due to NaOH. It comes out to .375 moles of NaOH.
I think you've misunderstood the question. CO2 is passed into a solution containing NaOH. The mass of this solution increases as a result of passing CO2 through it.
 
  • #3
Gokul43201 said:
I think you've misunderstood the question. CO2 is passed into a solution containing NaOH. The mass of this solution increases as a result of passing CO2 through it.


So the product CO2 is passed into a solution containing NaOH?

Or you are saying CO2 is in the reactants?

Could you possibly who what you think is happening with a chemical formula? Where did the extra 15g come from?
 
  • #4
Mass of the NaOH solution increased when the carbon dioxide was absorbed by this solution.

Carbon dioxide is produced in the first reaction (acid+carbonate) and consumed in the second reaction (separate solution of NaOH).

Hopefully it is clear now.
 
  • #5
Borek said:
Mass of the NaOH solution increased when the carbon dioxide was absorbed by this solution.

Carbon dioxide is produced in the first reaction (acid+carbonate) and consumed in the second reaction (separate solution of NaOH).

Hopefully it is clear now.

Borek
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Does this mean I was correct?

H2SO4 + MgCO3 -> MgSO4 + H2O + CO2
CO2 + NaOH -> NaHCO3

Hence
H2SO4 + MgCO3 -> MgSO4 + H2O + NaHCO3

So the answers at the back of the book were wrong?
 
  • #6
Forget about NaOH, it doesn't play any role here. NaOH solution part of the question is just a way of telling you what was the mass of the CO2 evolved. Insert this information into first reaction equation and you will be done.

Looks to me like 426 is a correct answer. EBAS rulez ;)


EBAS - equation balancer and stoichiometry calculator
 
  • #7
pivoxa15 said:
I don't understand. There is an increase in weight of 15g. Where did 15g come from if not from NaOH?
Increase in the weight of what?

(weight of NaOH solution after passing CO2) - (weight of NaOH before passing CO2) = 15g.

Clear now?
 
  • #8
Gokul43201 said:
I think you've misunderstood the question. CO2 is passed into a solution containing NaOH. The mass of this solution increases as a result of passing CO2 through it.

Gokul43201 said:
Increase in the weight of what?

(weight of NaOH solution after passing CO2) - (weight of NaOH before passing CO2) = 15g.

Clear now?

Borek said:
Forget about NaOH, it doesn't play any role here. NaOH solution part of the question is just a way of telling you what was the mass of the CO2 evolved. Insert this information into first reaction equation and you will be done.

Looks to me like 426 is a correct answer. EBAS rulez ;)

Borek
--
EBAS - equation balancer and stoichiometry calculator
www.pH-meter.info


Now it's clear. The solution in that final sentence in the question was pointing to the solution containing NaOH. The 15g was the CO2 from the first reaction. It too some time for this question but I finally understood. Thanks.
 

1. How do I calculate the volume of reactant H2SO4 for a reaction with MgCO3?

The volume of reactant H2SO4 can be calculated using the equation: V(H2SO4) = (n(MgCO3) x M(H2SO4) x V(H2O)) / (n(H2SO4) x M(MgCO3)), where n represents the number of moles and M represents the molar mass. The volume of water (V(H2O)) is typically negligible and can be disregarded in the calculation.

2. What is the balanced chemical equation for the reaction between H2SO4 and MgCO3?

The balanced chemical equation for this reaction is: H2SO4 + MgCO3 → MgSO4 + CO2 + H2O

3. How do I determine the number of moles of H2SO4 needed for the reaction?

The number of moles of H2SO4 needed can be calculated using the equation: n(H2SO4) = n(MgCO3) x M(MgCO3) / M(H2SO4), where M represents the molar mass.

4. What is the molar mass of H2SO4 and MgCO3?

The molar mass of H2SO4 is 98.08 g/mol and the molar mass of MgCO3 is 84.31 g/mol.

5. How do I convert the volume of reactant H2SO4 from liters to milliliters?

To convert from liters to milliliters, simply multiply the volume in liters by 1000. For example, if the volume of H2SO4 is 0.5 L, the volume in milliliters would be 0.5 L x 1000 = 500 mL.

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