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# 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?

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

### 4 Answers by Expert Tutors

David L. | Chemistry TutorChemistry Tutor
5.0 5.0 (23 lesson ratings) (23)
1
Magnesium hydroxide reacts with hydrochloric acid to form magnesium chloride and water.

Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O

This balanced equation, which is pretty typical of an acid-base reaction (acid + base → salt + water) is needed to figure out the mole ratio of acid to base.  The mole ratio is "1 mol Mg(OH)2 = 2 mol HCl".  You're given the volume (mL) and molarity (mol/L) of the acid, which can be used to find moles of acid.  You need to find the grams of base, so it's a good idea to write up a conversion plan.

It's often a good idea to start your plan with the answer's units.  To get grams of base, you'll need the moles of base.  You can get the moles of base from the balanced equation and the moles of acid.  You can get the the moles of acid if you know the volume in Liters and the molarity, which is mol/L.  Volume in L can be found from the volume in mL.

Plan: mL HCl → L HCl → mol HCl → mol Mg(OH)2 → grams Mg(OH)2

To do the last conversion, you'll need to calculate the molar mass of magnesium hydroxide.
• molar mass = 24.31 + (2 x 16.00) + (2 x 1.008) = 58.326 g/mol

Now the conversion can be put together and solved.  Don't round anything until the end, or the answer will probably be off by a bit. The molarity of the acid limits your final answer to two sig figs.

25 mL HCl x  (1 L / 1000 mL) x (0.10 mol HCl / 1 L HCl) x (1 mol Mg(OH)2 / 2 mol HCl)
x (58.326 g Mg(OH)2 / 1 mol Mg(OH)2)  = 0.073 grams Mg(OH)2

Yulia G. | Chem/biochem tutorChem/biochem tutor
5.0 5.0 (1 lesson ratings) (1)
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First, you need to write the reaction equation and balance it: Mg(OH)2 + 2HCl = MgCl2 + H2O.
From the equation you know that 1 mole of Mg(OH)2 can neutralize 2 moles of HCl.
Now you need to calculate how many moles of HCl needs to be neutralize. For this purpose you need to multiply concentration of the acid by its volume (convert in L): 0.1M (that is moles per L)*0.025 L (converted from 25 mL)=0.0025 moles. To neutralize this amount of HCl, 0.0025/2=0.00125 moles of Mg(OH)2 is required. To calculate the mass of Mg(OH)2 corresponding to this amount of moles, you need to know the molecular weight of Mg(OH)2: M(Mg)+2*M(O)+2*M(H)=24.3+2*16+2*1=57.3 g/mole. To calculate the mass you need to multiply the molecular weight by moles: 57.3*0.00125=0.0716 g or 71.6 mg. The answer is 71.6 mg of Mg(OH)2
Edmondo C. | Doctor of Chemistry. Chemistry, Science, Math, and Italian TutoringDoctor of Chemistry. Chemistry, Science,...
4.8 4.8 (35 lesson ratings) (35)
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Although there are several valid calculation procedures for this question (see other answers),
practicing chemists prefer simplify the calculations needed by introducing an operational concentration, normality(N), and the concept of gram equivalent( Eq) of a compound ( see my answer to the question of neutralization of a phosphoric acid solution with a sodium hydroxide solution for definitions and calculations). In this case we have a 0.1M solution of HCl that converts in a 0.1N solution. Hence 25ml of this solution contain 0.1x25 mEq of HCl. = 2.5 mEq.
By definition 2.5 mEq of HCl require 2.5 mEq of Mg(OH)2 for a quantitative neutralization.
Since the gram equivalent of Mg(OH)2 ( calculated as formula weight of MG(OH)2/2)
Is 58.3/2=29.15g , 2.5mEq correspond to
2.5 X 29.15 = 72.87mg

Regards.
Edmondo C.
Brad M. | STEM Specialist plus Business, Accounting, Investment & EditingSTEM Specialist plus Business, Accountin...
4.9 4.9 (231 lesson ratings) (231)
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Hey Brieonna -- neutrailizing acid usually means making water HOH ...

each M of HCl needs a M of OH- ... Mg(OH)2 provides a PAIR ... need 0.05M of Mg ...

each M of Mg(OH)2 has about 58g/mol ... need 1/20th or about 2.9g of Mg per L ...

to counter 25ml of 0.10M HCl, need 1/40th of the 3g or about 75mg of Mg ... Regards :)