[ainoko_hikaru]

I don't know if I'm on the right track. please help.
       A mixture of Na₂CO₃ and MgCO₃ of mass 7.63g is reacted with excess hydrochloric acid. The CO₂ gas generated occupies a volume of 1.67L at 1.24 atm and 26 degrees Celcius. From these data, calculate the percent composition by mass of Na₂CO₃ in the mixture.

I have already made a balanced equation:

Na₂CO₃ + MgCO₃ + 2HCl -> 2NaCl + MgCl₂ + 2CO₂ + 2H₂O

after that, I computed for the number of moles of CO₂ using the ideal gas equation and arrived with 0.0860 moles CO₂. I then calculated for the mass of carbon dioxide, then using this I calculated for the number of moles of Na₂CO₃ which turned out to be 4.29 x 10^-2. After computing for the number of moles I then computed for the mass of Na₂CO₃ which turned out to be 4.55 grams. My solution for percent composition involved using the computed mass of Na₂CO₃ divided by the given mass of Na₂CO₃ and MgCO₂ multiplied by a hundred resulting to 59.6%. Am I correct?

Thanks!

[sjb]

I'm not sure if you're right, but the approach seems wrong.

Essentially you have 2 things going on here

Na₂CO₃ + 2HCl -> 2CO₂ + 2NaCl + 2H₂O, and
Mg₂CO₃ + 2HCl -> 2CO₂ + MgCl₂ + 2H₂O.

So you have mass of the solid, which is equal to the mass of Na₂CO₃ + the mass of Mg₂CO₃, and the amount of CO₂, which comes from the decomposition.

Two equations, two unknowns.

What I think you have done is calculated the mass of Na₂CO₃ if there were an equimolar amount of Mg₂CO₃ and Na₂CO₃ present.

[enahs]

sjb is correct, your answer is wrong.

However, I do not like the two equation approach for this because it puts more emphasis on math and not thinking about what is going on behind it, in my opinion. But yes, math is required.

You have a mixture of two components, producing the same thing. Think about it at a molar level first.


1 mol Na₂CO₃ + 1 mol MgCO₃ -> 2 mol CO₂

You know the value for the total mols of CO2 produced (and I would use more significant figures in your calculations and do not round until the end, it is off a decent amount).

So you know

x mols Na₂CO₃ + y mols MgCO₃ -> 0.084 mols CO₂

Correct?
But you only know the mass, but how can you relate mass to mols?
Molecular Weight!

So, call one arbitrarily x and the other 7.63-x (because you know you have 7.63 grams total)

So now you have

   x g Na₂CO₃                      +         7.63 -x MgCO₃              =  0.084
 105.99 g Na₂CO₃ /mol                     84.32 g /  MgCO₃

Which the units cancel and leave mols + mols = 2 mols, which is what your balanced equation said.
You use the molecular weight to relate the mass to the amount of mols, as dictated by stoichmetry, directly into the balanced equation with the appropriate coefficient. It is no longer a system of equations and just a math problem, but the math is now displaying the chemical relationships.

Remember, balanced equation relates mols, not mass!

Now it is just the basic algebra and solving for x, and once you know x you know the other as well.

Hope that makes sense and helps.

[ainoko_hikaru]

oh, ok, thanks for pointing out my mistake. I'll try out the solution. Thanks for the help, enahs and sjb!