How much (in grams) calcium carbonate is needed in order to neutralize 100mL, 0.05 mole/L HCl solution?
First you need the balanced equation of calcium carbonate (CaCO3) reacting with HCl.
CaCO3 (s) + 2HCl (aq) --> H2O (l) + CO2 (g) + CaCl2 (aq)
Then, we need to see how many HCl present in the solution needs to be neutralized.
100 mL = 0.1 L to match units.
moles HCl = (0.1 L)(0.05 mole/L) = 0.005 mol HCl.
Then simple stoichiometry.
Hope that helps.
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