# 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?

Jun 27, 2017

$\text{0.073 g}$

#### 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.

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

You know that the hydrochloric acid solution contains $0.10$ moles of hydrochloric acid for every $\text{1 L} = {10}^{3}$ $\text{mL}$ 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

$0.00125 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Mg"("OH")_2))) * "58.32 g"/(color(red)(cancel(color(black)("moles Mg"("OH")_2)))) = color(darkgreen)(ul(color(black)("0.073 g}}}}$

The answer is rounded to two sig figs.