Calculate the volume of 0.100 M HCl required to neutralize 1.00 g of Ba(OH)₂ (molar mass = 171.3 g/mol).

Question

Quizplus · Accepted Answer

First, calculate the number of moles of Ba(OH)2:

1.00 g / 171.3 g/mol = 0.00584 mol Ba(OH)2

Since HCl and Ba(OH)2 react in a 1:2 ratio, we need twice as many moles of HCl:

0.00584 mol Ba(OH)2 × 2 mol HCl / 1 mol Ba(OH)2 = 0.0117 mol HCl

Finally, calculate the volume of 0.100 M HCl needed to contain this many moles of HCl:

0.0117 mol HCl ÷ 0.100 mol/L = 0.117 L = 117 mL

Therefore, the answer is B.

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