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

Question

A) 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.]


b)When a 0.6333-g sample of a compound of carbon, hydrogen and sulfur was burned in excess oxygen, 0.9272 g CO2, 0.3795 g H2O and 0.6749 g SO2 were obtained. In a separate experiment, it was found that, for fixed temperature and pressure, a 0.100 L sample of this compound takes 1.97 times longer to effuse than does a 0.100 L sample of Ar(g). What is the molecular formula? And draw possible structures for the molecule

Explanation / Answer

moles NaOH = 0.01981 L x 0.1026 = 0.002033 moles HCl = 0.108 L x 0.2045 = 0.02209 moles HCl used to titrate Mg(OH)2 = 0.02209 - 0.002033 = 0.02006 Mg(OH)2 + 2 HCl >> MgCl2 + 2 H2O moles Mg(OH)2 = 0.02006/2 =0.01003 Mass Mg(OH)2 = 0.01003 mol x 58.3178 g/mol = 0.5849 g % = 0.5849 x100 / 0.5899 = 99.16

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