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#?

Question

Henry Antrim · Accepted Answer

#"0.073 g"#

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.

Caleb Adams · Answer

undefined To neutralize 25 mL of 0.10 M HCl, you would need 0.25 grams of Mg(OH)2.