how many grams of CO2 are released in the combustion of one gram of CACO3?

Question

the combustion of CACO3 gives CaO + CO2
how many grams of CO2 are produced in the decompotition of 1 gram of CACO3?

Answer 1

here is one way to do it, (you don't always have to use moles):

[12 + ( 2 x 16)]/{40 + [12 + (3 x 16)]} x 1g = 0.44g

Answer 2

First you need to set up a balanced equation for the reaction of CaCO3 to give off CaO and CO2. Fortunately, the equation is already balanced and should look something like this...

CaCO3 --> CaO + CO2

You can tell by looking at the numercal coefficients in the balanced chemical equation how many moles of each are produced in this reaction. In this case it's 1:1:1. So for each mole of CaCO3 that reacts, one mole of CO2 will be produced.

Now you need to figure out how many moles of CaCO3 is present in one gram of the compound. Do this by diving by the molecular weight which is approximately 40+12+16+16+16 = 100.

1 gram/100 grams/mol = 0.01 moles

Now, reverting back to the balanced equation, you know that since the ratio is 1:1:1, 0.01 moles of CO2 must ALSO be produced. Now multiply this number by the molecular weight for CO2, (which is approximately 44 g/mol) and you get 0.44 grams.

Make sense? Good luck!

PS. You should try this again with a more complicated equation to make sure that you know the procedure. Because the equation was already balanced for you, you didn't have to do any number manipulations. But if you had one mole of CaCO3 per TWO moles of CO2, you would have to multiply that number (0.01 mol) by TWO because for every mole of CaCO3 that reacts, TWO moles of CO2 are produced. Just always make sure that your equation is balanced before proceeding with the calculations or else all you numbers will be off...

Regards,
Mrs. Mustang

Answer 3

1 gram CaCO3 x 1mol/100.1 = 0.010 mol CaCO3.

0.010 mol CaCO3 = 0.010 mol CO2 x 44.01g/mol = .4401 grams CO2