A 0.2527-g sample of a mixture of Mg(OH)2 and Al(OH)3 was treated with 50.00 mL

Question

A 0.2527-g sample of a mixture of Mg(OH)2 and Al(OH)3 was treated with 50.00 mL of 0.225 mol L 1 HCl(aq). The resulting solution required 13.73 mL of 0.176 mol L 1 NaOH(aq) to neutralize the excess, unreacted HCl.
What was the percentage by mass of Mg(OH)2 in the original sample? [Hint: When Mg(OH)2 or Al(OH)3 reacts with HCl(aq), the products are a chloride salt, water and CO2(g). Assume that all of the CO2(g) leaves the solution without reacting.]

Explanation / Answer

From the neutralization data, you can calculate the moles of NaOH used in the neutralization, which equals the moles of unreacted HCl (mol = M * L). Subtract this from the moles of of HCl present initially (25mL, 0.953M) and you have the moles of acid that were needed to neutralize the OH- in the Mg(OH)2 tablet. Now you have moles of OH- in the tablet, which you can convert to moles of Mg(OH)2 in the tablet. Remember that there are two moles of OH- per one mole of Mg(OH)2: Mg(OH)2(aq) + 2HCl(aq) ---> 2H2O(l) + MgCl2(aq) And now you just have to use the molar mass of Mg(OH)2 to convert moles of Mg(OH)2 in the tablet to grams of Mg(OH)2 in the tablet. The mass percent will be (g Mg(OH)2 / g tablet)(100%).

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