Asked by Christina
Calcium carbonate reacts with hydrochloric acid according to the following reaction:
CaCO3 (s) + 2 HCl (aq) --> CaCl2 (aq) + H2O (l) + CO2 (g)
How many grams of calcium carbonate would be required to neutralize 50.0 cm3 of 8.00 M HCl?
CaCO3 (s) + 2 HCl (aq) --> CaCl2 (aq) + H2O (l) + CO2 (g)
How many grams of calcium carbonate would be required to neutralize 50.0 cm3 of 8.00 M HCl?
Answers
DrBob222
Convert HCl to moles. moles = M x L.
Using the coefficients in the balanced equation, convert moles HCl to moles CaCO3.
Now convert mols CaCO3 to grams. g = moles x molar mass.
Using the coefficients in the balanced equation, convert moles HCl to moles CaCO3.
Now convert mols CaCO3 to grams. g = moles x molar mass.