When surface water dissolves carbon dioxide, carbonic acid (H2CO3) is formed. When the water moves underground through limestone formations, the limestone dissolves and caves are sometimes produced.
CaCO3(s) + H2CO3(aq) Ca(HCO3)2(aq)
What mass of limestone must have dissolved if 1.98 1010 kg of calcium hydrogen carbonate was produced?
To calculate the mass of limestone that must have dissolved, we need to use stoichiometry. From the balanced chemical equation:
CaCO3(s) + H2CO3(aq) -> Ca(HCO3)2(aq)
We can see that one mole of calcium carbonate (CaCO3) reacts with one mole of carbonic acid (H2CO3) to produce one mole of calcium hydrogen carbonate (Ca(HCO3)2).
First, let's calculate the molar mass of calcium hydrogen carbonate (Ca(HCO3)2):
Ca(HCO3)2:
Calcium (Ca) atomic mass: 40.08 g/mol
Hydrogen (H) atomic mass: 1.01 g/mol
Carbon (C) atomic mass: 12.01 g/mol
Oxygen (O) atomic mass: 16.00 g/mol
Total molar mass: 40.08 + 2*(1.01) + 2*(12.01) + 4*(16.00) = 162.12 g/mol
Now, we can set up a proportion to find the mass of limestone that must have dissolved:
(1.98 x 10^10 kg Ca(HCO3)2) * (1 mol CaCO3 / 162.12 g Ca(HCO3)2) * (1000 g / 1 kg) = x mol CaCO3
Simplifying the expression:
x = (1.98 x 10^10 kg * 1 mol CaCO3 * 1000 g) / (162.12 g Ca(HCO3)2)
Calculating the result:
x = 1.22 x 10^11 mol CaCO3
To convert this to mass, we can use the molar mass of calcium carbonate (CaCO3):
CaCO3:
Calcium (Ca) atomic mass: 40.08 g/mol
Carbon (C) atomic mass: 12.01 g/mol
Oxygen (O) atomic mass: 16.00 g/mol
Total molar mass: 40.08 + 12.01 + 3*(16.00) = 100.09 g/mol
Finally, we calculate the mass of limestone:
Mass = x mol CaCO3 * 100.09 g/mol = (1.22 x 10^11 mol * 100.09 g/mol) = 1.22 x 10^13 g
So, approximately 1.22 x 10^13 grams of limestone must have dissolved to produce 1.98 x 10^10 kg of calcium hydrogen carbonate.