Question #179657

What volume in ml of 1.5 *M* HCl is needed to completely react with 92 grams of the CaCO_{3}?

Expert's answer

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

CaCO_{3 }+ HCL = CaCl_{2 }+ H_{2}O + CO_{2}

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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