Question

How many milliliters of 0.0197m ca(oh)2 would be required to exactly neutralize 136mm of 0.258m hcl?

Answer 1

number of moles HC

Volume HCl in liters : 136 mL / 1000 => 0.136 L

0.258 x 0.136 => 0.035088 moles of HCl

Mole ratio:

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

2 moles HCl ------------------------ 1 mole Ca(OH)2
0.035088 moles HCl ------------- moles Ca(OH)2

moles Ca(OH)2 = 0.035088 x 1 / 2

moles Ca(OH)2 = 0.035088 / 2

= 0.017544 moles of Ca(OH)2

M = n / V

0.0197 = 0.017544 / V

V = 0.017544 / 0.0197

V = 0.890 L

0.890 in mL : 0.890 x 1000 => 890 mL

hope this helps!