Question

Question #6e34d

Answer 1

`"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

`"Mg"("OH")_ (2(s)) + 2"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))`

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 color(red)(cancel(color(black)("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.