Chem help needed?

Question

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 man mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H3PO4 solution?

Answer 1

a) 25 ml HCl at 0.01 M is

25 ml * 0.1 Moles HCl/1000 ml = 0.0025 moles


Mg(OH)2 MW = 58.33 g/mole

Mg(OH)2(aq) + 2HCl(aq) --> MgCl2(aq) + 2H2O(l)

each HCl needs 1/2 mole Mg(OH)2

so we need 0.0025/2 moles Mg(OH)2 = 0.00125 mole Mg(OH)2

0.0125 moles Mg(OH)2 = 0.00125 moles * 58.33 g/mole =

= 0.073 g = 73 mg Mg(OH)2

b) 15 ml * 0.2 mole H3PO4/1000ml = 0.003 mole

3NaOH(aq) + H3PO4(aq) --> Na3PO4(aq) + 3H2O(l)

3 moles NaOH will be needed to neutralized
the H3PO4 =

3 * 0.003 = 0.009 moles

0.009 mole = xml * 0.10moleNaOH/1000ml

xml = 90 ml NaOH

Answer 2

Atomic weights: Mg=24 O=16 H=1 Mg(OH)2=58 Cl=35.5 HCl=36.5 P=31 H3PO4=98

Mg(OH)2 + 2HCl ===> MgCl2 + 2H2O

Let stomach acid be called SA.

25mLSA x 0.10molHCl/1000mLSA x 1molMg(OH)2/2molHCl x 58gMg(OH)2/1molMg(OH)2 = 0.0725g Mg(OH)2

3NaOH + H3PO4 ===> Na3PO4 + 3H2O

Let the phosphoric acid solution be called PS. Let the NaOH solution be called NS.

15mLPS x 0.20molH3PO4/1000mLPS x 3molNaOH/1molH3PO4 x 1000mLNS/0.10molNaOH = 90 mL NaOH solution