Neutrilization of Acids and Bases

[the prodogy]

Need help figuring out problem... again:

How many milliliters of 0.60 M HCl are required to neutralize 3.0 grams of CaCO3?

A. 50 mL
B. 100 mL
C. 200 mL
D. 300 mL

I looked at the answer and how they come up with it, I'm not sure how they're coming up with some of the numbers.

Thanks

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[loveoforganic]

2HCl + CaCO3 => H2CO3 + H2O + CO2

(grams CaCO3)*(1 mol / MW CaCO3)*(2 mol HCl / 1 mol CaCO3)*(1 L HCl / 0.60 mol HCl)*(1000 mL / L)

(3.0)*(1 / 100.09)*(2)*(1 / 0.60)*(1000) = 99.9 mL