1)How many moles of HCl are in 67 mL of 0.11 M HCl? 2)What mass of calcium carbo

Question

1)How many moles of HCl are in 67 mL of 0.11 M HCl?

2)What mass of calcium carbonate is needed for complete reaction with the HCI in the previous question

HCl in (a)?Calcium carbonate reacts with HCl according to the following equation:
2HCl(aq)+CaCO3(s)CaCl2(aq)+H2O(l)+CO2(g)

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Explanation / Answer

1.

use the formula

Moles = Molarity of the solution x volume of the solution in liters

given volume is 67 mL = 0.067 L

moles = 0.11M x 0.067 L = 0.00737 mol

2.

lets write the balanced equation

CaCO3 (s) + 2HCl (aq) ---> CaCl2 + H2O (l) + CO2 (g)

from the balanced equation

1 mol of HCl required half mole of CaCO3; accordingly

0.00737 mol of HCl required 0.00737 mol / 2 = 0.003685 mol CaCO3

mass of CaCO3 = moles x molar mass = 0.003685 mol x 100.0869 g/mol = 0.37 g

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