Question

How many grams of `Mg(OH)_2` will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M `HCl`?

Answer 1

`"0.073 g"`

Explanation:

The trick here is to keep in mind that you need `2` moles of hydrochloric acid in order to neutralize `1` mole of magnesium hydroxide.

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

You know that the hydrochloric acid solution contains `0.10` moles of hydrochloric acid for every `"1 L" = 10^3` `"mL"` of solution, so you can say that your sample contains

`25 color(red)(cancel(color(black)("mL"))) * "0.10 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))) = "0.0025 moles HCl"`

This means that a complete neutralization will require

`0.0025 color(red)(cancel(color(black)("moles HCl"))) * ("1 mole Mg"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.00125 moles Mg"("OH")_2`

Finally, to convert this to grams, use the molar mass of the compound

`0.00125 color(red)(cancel(color(black)("moles Mg"("OH")_2))) * "58.32 g"/(color(red)(cancel(color(black)("moles Mg"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.073 g")))`

The answer is rounded to two sig figs.