# Question 6e34d

Jul 26, 2017

"0.66 g Mg"("OH")_2

#### Explanation:

You know that magnesium hydroxide and hydrochloric acid neutralize each other to produce water and aqueous magnesium chloride

${\text{Mg"("OH")_ (2(s)) + 2"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

so start by using the molarity and volume of the hydrochloric acid solution to determine how many moles are present in the sample.

225 color(red)(cancel(color(black)("mL solution"))) * "0.10 moles HCl"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0225 moles HCl"

Now, you know that the reaction consumes $2$ moles of hydrochloric acid for every $1$ mole of magnesium hydroxide that takes part in the reaction.

Use this $1 : 2$ mole ratio to find the number of moles of magnesium hydroxide needed

0.0225 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole Mg"("OH")_2)/(2color(red)(cancel(color(black)("moles HCl")))) = "0.01125 moles Mg"("OH")_2#

Finally, to convert this to grams, use the molar mass of magnesium hydroxide

$0.01125 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Mg"("OH")_2))) * "58.32 g"/(1color(red)(cancel(color(black)("mole Mg"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.66 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the hydrochloric acid solution.