Question

The volume of 0.300 mol/L HCl that can be completely neutralized by 2.00 g of Ca(OH)2 in a stomach antacid tablet is? Please help me show the work.

Answer 1

Here M of HCl = 0.300 mol/L

Measn 0.300 mol of HCl is present in HCl.

First write the reaction equation;

Ca(OH)2 + 2HCl --> CaCl2 + 2H2O

Now calcualte the number of moles of Ca(OH)2 which will use to neutralized:

0.300 mol of HCl * 1 mol of Ca(OH)2 / 2 mol of HCl

= 0.15 mol Ca(OH)2

Now calcualte the number of moles of Ca(OH)2 as followS:

2.00 g / 74.0927 g/ mol = 0.02699 mol

But here the moles of Ca(OH)2 is less than required mole sof Ca(OH)2 hence it cannot be neutrilized.