Answer to Question #179657 in General Chemistry for Enci Peng

Question #179657

What volume in ml of 1.5 M HCl is needed to completely react with 92 grams of the CaCO3?


1
Expert's answer
2021-04-13T02:08:35-0400

first calculate the moles of calcium carbonate

Moles = "\\frac{Mass}{molar mass} =" "\\frac{92}{100.0869}=0.917" moles

the equation of reaction gives a mole ratio between calcium carbonate and hydrochloric acid as 1:1

CaCO3 + HCL = CaCl2 + H2O + CO2

therefore to find the moles used for HCl=(0.917×1)= 0.917moles

to find this vollume= "\\frac {mol\/l}{Molarity} =" (0.917 × 1000) ÷ 1.5 = 611. 6 ml

Volume = 611.6 ml


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