Question

The relative molecular mass of calcium carbonate (CaCO3) is 100. What is the minimum volume of 2.0 M hydrochloric acid (HCl) that would be needed to completely react with 2.0 g of calcium carbonate?

Answer 1

CaCO3 + 2HCl > CaCl3 + CO2 + H2O

2grams/100= .02 moles
so you need .04 moles HCl

molesHCl=volume*Molarity
volume = .04/2= .02 liters or 20 ml

Answer 2

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Answer 3

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Answer 4

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