Calculating Mass

Question

Milk of magnesia has 400 mg of  Mg (OH)_2 per teaspoon. Calculate the mass of stomach acid that can be neutralized by 1.70 teaspoon of milk of magnesia, assuming that the stomach acid is 0.50% HCI by mass. 

Express your answer with appropriate units.

Answer 1

Mg(OH)₂ + 2 HCl ==> Mg(Cl)₂ + 2 H₂O ... balanced equation

1.7 tsp........? grams

1.7 tsp x 400 mg/tsp = 680 mg Mg(OH)2

moles Mg(OH)2 = 680 mg x 1 g/1000 mg x 1 mol Mg(OH)2/58.3 g = 0.01166 moles Mg(OH)2 used

moles HCl neutralized = 0.01166 mol Mg(OH)2 x 2 mol HCl/mol Mg(OH)2 = 0.02332 moles HCl

mass HCl neutralized = 0.02332 mol HCl x 36.5 g /mol = 0.851g = 851 mg HCl neutralized

The fact that it is 0.5% by mass does not enter into the calculations, unless you want to solve for the VOLUME of stomach acid.

Answer 2

Mg(OH)₂ + 2 HCl - Mg(Cl)₂ + 2 H₂O. That is, 1 mole of Mg(OH)₂ wil neutralize 2 moles of HCl.

Rx 1 teaspoon of Milk of Magnesia has 400 mg of Mg(OH)2.

So 1.7 teaspoonful will have 680 mg of Mg(OH)₂ [1.7 x 400 = 680 ]

----------------------------------------------------------------------------------------------------------------------

Since 1 mole of Mg(OH)₂ will neutralize 2 moles of HCl,

680 mg of Mg(OH)₂ will neutralize 1360 mg of HCl.

---------------------------------------------------------------------------------------------------------------------------

Strength of HCl in the stomach as per the question is 0.5%, which means 0.5 grams of HCl/100 mL.

500 mg (0.5 g) in 100 mL

1360 mg of HCl will be in 1360 x 100 /500 = 272 mL.

Mass of HCl that can be neutralized by 1.7 teaspoon of Milk of Magnesia is 1360 mg in 272 mL