How many grams of Mg(OH)_2Mg(OH)2 will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M HClHCl?

1 Answer
Stefan V.
Jun 27, 2017

"0.073 g"0.073 g

Explanation:

The trick here is to keep in mind that you need 22 moles of hydrochloric acid in order to neutralize 11 mole of magnesium hydroxide.

"Mg"("OH")_ (2(s)) + color(red)(2)"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + 2"H"_ 2"O"_ ((l))Mg(OH)2(s)+2HCl(aq)→MgCl2(aq)+2H2O(l)

You know that the hydrochloric acid solution contains 0.100.10 moles of hydrochloric acid for every "1 L" = 10^31 L=103 "mL"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.

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