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How much (in grams) calcium carbonate is needed in order to neutralize 100mL, 0.

Chemistry
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How much (in grams) calcium carbonate is needed in order to neutralize 100mL, 0.05 mole/L HCl solution?

Nov 24th, 2014

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.

0.005 mol HCl  *1 mol CaCO3  *100.09 g CaCO3  =0.25 g CaCO3                                                                  

2 mol HCl1 mol CaCO3

The answer is 0.25 g CaCO3.

Hope that helps.

Nov 25th, 2014

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