This 5 point homework problem is due on Friday April 21 at 11 AM. Digul circled

Question

This 5 point homework problem is due on Friday April 21 at 11 AM. Digul circled R is an antacid whose active ingredients are solid aluminum hydroxide, and solid magnesium hydroxide. Stomach acid is primarily an aqueous solution of 0.12 M hydrochloric acid. How many mL of stomach acid will neutralized by the magnesium hydroxide in two 1.25 gram tablet of Digul, assuming that the tablets are 74%magnesium hydroxide? for full credit including a labeled and balanced chemical equation. No Chit! for full credit including a labeled and balanced chemical equation. No Chit! Work submitted on any other size or color will not be scored!

Explanation / Answer

Ans. Part 1: Mass of Mg(OH)2 in one tablet = 74 % (w/w) of 1.25 g = 0.925 g

Total mass of Mg(OH)2 in two tablets = 2 tablet x [ Mass of Mg(OH)2 / tablet ]

                                                            = 1.85 g

Total moles of Mg(OH)2 = Mass/ Molar mass

                                                = 1.85 g / (58.31968 g/mol)

                                                = 0.031722 mol

Part 2: Balanced reaction: Mg(OH)2(aq) + 2 HCl(aq) ----------> MgCl2(aq) + 2H2O(aq)

Stoichiometry: 1 mol Mg(OH)2 requires 2 mol HCl for complete neuralization.

At equivalence point, the total number of moles of OH- ions must be equal to that of H+.

Since, 1 mol Mg(OH)2 gives 2 OH-,

the total number of moles of OH- = 2 x moles of Mg(OH)2

                                                                                = 2 x 0.031722 mol

                                                            = 0.063444 mol

So, at equivalence point, moles of H+ = Moles of OH- = 0.063444 mol

Therefore, the number of moles of H+ from gastric juice = 0.063444 mol

Part 3: 1 mol HCl gives 1 mol H+. Therefore, number of moles of HCl at equivalence point is equal to 0.063444 mol.

Now,

Number of moles of HCl = Molarity of HCl solution x Vol. of solution in litres

            Or, 0.063444 mol = 0.12 M x V L

            Or, 0.063444 mol = (0.12 mol/ L) x V L = 0.12 mol x V

            Or, V = 0.063444 mol / 0.12 mol = 0.5287

Therefore, required volume of HCl solution = V litres = 0.5287 L = 528.7 mL

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.