Limiting reactants product of moles

[amesbruin1L]

How many moles of CO2(g) are produced when 1 kg of CaCO3(s) is used to neutralize an acid spill? The equation for the reaction at 1 atm and 25 degrees C is:

CaCO3(s) + H2SO4(aq) ---> CaSO4(s) + CO2(g) + H2O(l)


I got really stumped while working out this problem and can't seem to get past it. I was wondering if anyone could help?

[Emily Ng_4C]

You can convert the mass of CaCO3 to moles and then use the mole to mole ration from the chemical equation to find the moles of CO2.

[amesbruin1L]

so then it would be 10 moles?

[MaanasO 1A]

Hi Ames!

So the molar mass of calcium carbonate (CaCO3) is 100.0869 g/mol. Multiplying 1 kg by 1000 gives you 1000g of CaCO3.

Converting from mass to moles means 1000 g CaCO3 * 1 mol CaCO3/100.1 g CaCO3 = 10 mol CaCO3.

To answer this question, we can assume that CaCO3 is the limiting reactant and we have sulfuric acid (H2SO4) in excess.

Based on the reaction, we know that the mole ratio between CaCO3 and CO2 is 1:1. Using dimensional analysis, we can do the following:
10 mol CaCO3 * 1 mol CO2/1 mol CaCO3 = 10 mol CO2. So the final answer is 10 moles of carbon dioxide.

Hope that helps!

[Duby3L]

So first you would want to calculate the moles of reactants. Once you have the moles you can use the molar ratio from the balanced equation to see how many moles are required and figure which is your excess and limiting. Once you know that you can do a mole to mole ratio and mole to mass ratio.

Hope this helps a bit!