Question

What volume of 0.128 M HCl is needed to neutralize 2.87 G of Mg(OH)2?

Answer 1

V= .768L= 76.8ml HCL

Answer 2

0.769 dm³ of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)₂

First, we will write a balanced chemical equation for the reaction between HCl and Mg(OH)₂

2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O

This means 2 moles of HCl is required to neutralize 1 mole of Mg(OH)₂.

First, we will determine the number of moles present in 2.87 g of Mg(OH)₂

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass = 2.87 g

From the formula,

$Number \ of \ moles = \frac{Mass}{Molar \ mass}$

$Number \ of \ moles = \frac{2.87 g}{58.3197 \ g/mol}$

$Number \ of \ moles = 0.04921 \ moles$

Since 2 moles of HCl is required to neutralize 1 mole of Mg(OH)₂

Then, 2 × 0.04921 moles of HCl will be required to neutralize 0.04921 moles of Mg(OH)₂

∴ Number of moles of HCl required to neutralize the Mg(OH)₂ = 2 × 0.04921 moles

= 0.09842 moles

∴ 0.09842 moles of HCl is required to neutralize the Mg(OH)₂

Now, we will determine this in volume

Concentration of the HCl = 0.128 M

Number of moles = 0.09842 moles

Volume =?

From the formula,

Number of moles = Concentration × Volume

Volume = Number of moles ÷ Concentration

Volume = 0.09842 moles ÷ 0.128 M

Volume = 0.76890625 dm³

Volume ≅ 0.769 dm³

Hence, 0.769 dm³ of 0.128 M HCl is needed to neutralize 2.87 g of Mg(OH)₂

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