CHEM 182 Experimeos ve 4. Use the reaction belkow to answer the following questi

Question

CHEM 182 Experimeos ve 4. Use the reaction belkow to answer the following questions How may moles of stomach acid would he neutralized by one tablet of Maalox that contains 600 mg of cakciam carbonane? b. Assuming the volume of the stomach to be 1.01 with a pH of I, what will be the new pi of 500 mg of calcium carbonate? 5. Each 5-tl teaspoonful of Extra Strength Maslox Plus contains 450 mg of magnesiunm hydroxide and 500 mg of alumimum hydroxide. How many moles of hydronium ion are nestralized by one teaspoonful of Extra Strength Maalox Plus?

Explanation / Answer

Given

Mass of CaCO3 = 600 mg = 0.6 g

molar mass of CaCO3 = 100 g/mol

No. of moles of CaCO3 = Mass * molar mass = 0.6 g / 100 g/mol = 0.006 moles

according to given equation

CaCO3 + 2 H3O+ ---> Ca2+ (aq) + CO2 (g) + 3 H2O (l)

1 mole 2 mole

each mole CaCO3 requires 2 moles of H3O+ ( i.e acid)

so 0.006 moles of CaCO3 requires 0.012 moles of acid

so 600 mg of tablet requires 0.012 moles of stomach acid Answer 4(a)

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Given

pH = 1

- log ([H3O+]) = 1

[H3O+] = 0.1 mol/L

Given volume = 1 L

No. of moles of H3O+ present = 0.1 mol/L * 1 L = 0.1 moles

Mass of CaCO3 = 500 mg = 0.5 g

molar mass of CaCO3 = 100 g/mol

No. of moles of CaCO3 = Mass * molar mass = 0.5 g / 100 g/mol = 0.005 moles

according to given equation

CaCO3 + 2 H3O+ ---> Ca2+ (aq) + CO2 (g) + 3 H2O (l)

1 mole 2 mole

each mole CaCO3 requires 2 moles of H3O+ ( i.e acid)

so 0.005 moles of CaCO3 requires 0.010 moles of acid

so 500 mg of tablet requires 0.010 moles of stomach acid   

so amount of H3O+ remaining = 0.10 moles - 0.010 moles = 0.09 moles

[H3O+] = 0.09 moles / 1 L = 0.09 mol/L

pH = - log ([H3O+]) = - log(0.09) = 1.046

Answer pH = 1.046

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