Respuesta :
Answer: The volume of magnesium hydroxide needed to neutralize the given amount of HCl is 7.29 mL
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of HCl = 0.500 M
Volume of solution = 40 mL
Putting values in above equation, we get:
[tex]0.500M=\frac{\text{Moles of HCl}\times 1000}{40mL}\\\\\text{Moles of HCl}=0.02mol[/tex]
The chemical equation for the reaction of magnesium hydroxide and hydrochloric acid follows:
[tex]Mg(OH)_2+2HCl\rightarrow MgCl_2+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of HCl reacts with 1 mole of magnesium hydroxide.
So, 0.02 moles of HCl will react with = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of magnesium hydroxide
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of magnesium hydroxide = 58.32 g/mol
Moles of magnesium hydroxide = 0.01 moles
Putting values in equation 1, we get:
[tex]0.01mol=\frac{\text{Mass of magnesium hydroxide}}{58.32g/mol}\\\\\text{Mass of magnesium hydroxide}=(0.01mol\times 58.32g/mol)=0.5832g[/tex]
We are given:
Mass of magnesium hydroxide = 400 mg = 0.4 g (Conversion factor: 1 g = 1000 mg)
To calculate the volume of magnesium hydroxide needed to neutralize, we use unitary method:
For 0.4 g of magnesium hydroxide, the volume occupied is 5.00 mL
So, for 0.5832 g of magnesium hydroxide, the volume occupied will be = [tex]\frac{5.00}{0.4}\times 0.5832=7.29mL[/tex]
Hence, the volume of magnesium hydroxide needed to neutralize the given amount of HCl is 7.29 mL
The volume, in mL, of milk of magnesia that would be needed to neutralize 40.0 mL of 0.500 M HCl will be 7.29 mL
Stoichiometric calculations
From the equation of the reaction of magnesium hydroxide and HCl:
Mg(OH)2 + 2HCl ----------> MgCl2 + 2H2O
Mole ratio of Mg(OH)2 and HCl = 1:2
Mole of 40.0 mL 0.500 M HCl = 0.5 x 40/1000
= 0.02 moles
Equivalent mole of Mg(OH)2 = 0.02/2 = 0.01 moles
Mass of 0.01 moles Mg(OH)2 = 0.01 x 58.3
= 0.583 grams or 583 mg
Since 5.00 mL of milk of magnesia contains 400 mg , how many mL will contain 583 mg?
= 5 x 583/400
= 7.29 mL
More on stoichiometric calculation can be found here: https://brainly.com/question/8062886
