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27 January, 08:29

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

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  1. 27 January, 11:50
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    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
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