How many milliliters of 0.564 m hcl are required to react with 6.03 grams of caco3 ?

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asked Mar 14, 2019 54.8k views

1 Answer

Mukyuu · Answer 1 · 2019-03-18T16:28:16+0000

First write up a balanced equation
CaCO3 + 2 HCl = CO2 + CaCl2 + H2O

Find how many moles of CaCO3 you are dealing with
6.03 g x 1 mol / 100 g = .0603 moles

However, you need twice as much HCl
.0603 x 2 moles / .564 M = .21 L

How many milliliters of 0.564 m hcl are required to react with 6.03 grams of caco3 ?

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