Question

3. 0.050 mol of Ca(OH)2 are combined with 0.080 mol of HCl according to the following equation:
Ca(OH)2(aq) + 2HCl(aq) +CaCl2(aq) + 2H2O(1)
a. How many moles of HCl are required to neutralize all 0.050 mol
of Ca(OH)2?

Answer 1

The number of mole of HCl required to neutralize all 0.05 mole of Ca(OH)₂ is 0.1 mole

Balanced equation

Ca(OH)₂ + 2HCl —> CaCl₂ + 2H₂O

From the balanced equation above,

1 mole of Ca(OH)₂ required 2 moles of HCl

How to determine the mole of HCl required

From the balanced equation above,

1 mole of Ca(OH)₂ required 2 moles of HCl

Therefore,

0.05 mole of Ca(OH)₂ will require = 0.05 × 2 = 0.1 mole of HCl

Thus, 0.1 mole of HCl is needed for the reaction

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