Question

What mass of magnesium hydroxide is required to neutralize $ 125ml $ of $ 0.136M $ of $ HCl $ solution?
Molar mass of $ Mg{\left( {OH} \right)_2} = 58.33gmo{l^{ - 1}} $
(A) $ 0.248g $
(B) $ 0.496g $
(C) $ 0.992g $
(D) $ 1.98g $

Answer 1

Hint: Magnesium hydroxide and hydrochloric acid react with each other leads to the formation of magnesium chloride along with water. From the given values, moles of $ HCl $ can be determined. From these moles, the moles of magnesium hydroxide can be determined. From the obtained moles and molar mass, the mass of magnesium hydroxide can be calculated.

Complete Step By Step Answer:
Magnesium hydroxide is a compound with a molecular formula of $ Mg{\left( {OH} \right)_2} $ , it is a base. Hydrochloric acid is a strong acid with a molecular formula of $ HCl $ , it is an acid.
Strong acid and a strong base react with each other leads to the formation of an ionic salt and water. Here, magnesium hydroxide and hydrochloric acid react to form magnesium chloride and water.
The balanced chemical equation involved is as follows:
 $ Mg{\left( {OH} \right)_2} + 2HCl \to MgC{l_2} + 2{H_2}O $
Given volume of $ HCl $ is $ 125ml $ and the molar concentration of $ HCl $ is $ 0.136M $
Molar concentration is the ratio of the number of moles to the volume of solution in liters.
 $ 0.136M = \dfrac{{n \times 1000}}{{125}} $
By simplification, number of moles of $ HCl $ is $ 0.017 $
As two moles were required to neutralize one mole of magnesium hydroxide, the number of moles of magnesium hydroxide will be $ \dfrac{{0.017}}{2} = 0.0085 $ moles.
Given that molar mass of $ Mg{\left( {OH} \right)_2} = 58.33gmo{l^{ - 1}} $
Thus, the mass of $ Mg{\left( {OH} \right)_2} $ will be $ 0.0085mol \times 58.33gmo{l^{ - 1}} = 0.496g $
 $ 0.496 $ grams of magnesium hydroxide is required to neutralize $ 125ml $ of $ 0.136M $ of $ HCl $ solution

Note:
Based on the balanced chemical equation, moles of one reactant and another reactant can be calculated. A balanced equation must consist of the same number of atoms on both sides of the reaction. While calculating the molarity, the volume must be in liters, if it is in milliliters, multiply the obtained molarity by $ 1000 $ as one liter is equal to $ 1000ml $ .