Question

For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to neutralize? For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to neutralize 50.0 mL of 0.500 M HCl? Express your answer in mL.

Answer 1

Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O 

(0.0500 L) x (0.500 mol/L HCl) x (1 mol Mg(OH)2 / 2 mol HCl) x (58.3197 g Mg(OH)2/mol) = 0.729 g = 729 mg Mg(OH)2 

(729 mg Mg(OH)2) x (5.00 mL / 400 mg) = 9.11 mL milk of magnesia

Answer 2

Answer:

9.0625 mL of milk of magnesia are needed to neutralize 50.0 mL of 0.500 M HCl

Explanation:

Milk of magnesia is a compound that mainly contains magnesium hydroxide Mg (OH) 2. The balanced neutralization reaction with hydrochloric acid HCl is:

Mg(OH)₂ + 2 HCL ⇒ MgCl₂ + 2 H₂O

You need to neutralize 50.0 mL of 0.500 M HCl. Recalling that the molarity M indicates the amount of moles present in 1 liter of solution and that 1L = 1000 ml, it is possible to determine the moles that I need to neutralize as follows:

$moles of HCl=0.05 L*0.5\frac{mol}{L}$

moles of HCl=0.025  

Now, observing the stoichiometry of the reaction in the neutralization reaction it is possible to know the amount of moles of magnesium hydroxide necessary to neutralize 0.025 moles of hydrochloric acid, using again rule of three: if for 2 moles of hydrochloric acid to react, they are necessary 1 mole of magnesium hydroxide, how many moles of hydroxide are needed to neutralize 0.025 moles of hydrochloric acid?

$moles of Mg(OH)2=\frac{0.025molesofHCl*1molofMg(OH)2}{2molesofHCl}$

moles of Mg(OH)2=0.0125

The mass of Mg (OH) 2 is possible to know by knowing the mass of each element that composes it:

mass of Mg(OH)2= 24 + (16+1)*2

mass of Mg(OH)2= 58 g/mol

So, if one mole of magnesium hydroxide represents 58 grams of the compound, how many grams have 0.0125 moles of magnesium hydroxide?

$mass of magnesium hydroxide=\frac{0.0125 moles*58 grams}{1mol}$

mass of magnesium hydroxide= 0.725 g

Taking into account that 1 g = 1000 mg, so 0.725 g=725 mg

It is now possible to reapply a rule of three: if for every 5.00 mL of milk of magnesia there are 400 mg of magnesium hydroxide, then how many ml of milk of magnesia contain 725 mg of magnesium hydroxide?

$ml of milk of magnesia=\frac{725 mg*5 ml}{400 mg}$

ml of milk of magnesia=9.0625 ml

So, 9.0625 mL of milk of magnesia are needed to neutralize 50.0 mL of 0.500 M HCl