Correct Answer:<\/strong> Explanation:<\/strong> Using the given molar mass of aluminum sulfate (342.17 g\/mol<\/strong>), the number of moles in 25.0 g of Al\u2082(SO\u2084)\u2083 is calculated:Moles of Al2(SO4)3=25.0 g342.17 g\/mol\u22480.07307 mol\\text{Moles of Al}_2(\\text{SO}_4)_3 = \\frac{25.0\\ \\text{g}}{342.17\\ \\text{g\/mol}} \\approx 0.07307\\ \\text{mol}Moles of Al2\u200b(SO4\u200b)3\u200b=342.17 g\/mol25.0 g\u200b\u22480.07307 mol<\/p>\n\n\n\n Since each mole of aluminum sulfate contains 2 moles of aluminum, multiply the result:0.07307 mol\u00d72=0.14614 mol of Al0.07307\\ \\text{mol} \\times 2 = 0.14614\\ \\text{mol of Al}0.07307 mol\u00d72=0.14614 mol of Al<\/p>\n\n\n\n The molar mass of aluminum (Al) is 26.98 g\/mol<\/strong>. To find the mass of aluminum, multiply the number of moles by the atomic mass:0.14614 mol\u00d726.98 g\/mol\u22483.943 g0.14614\\ \\text{mol} \\times 26.98\\ \\text{g\/mol} \\approx 3.943\\ \\text{g}0.14614 mol\u00d726.98 g\/mol\u22483.943 g<\/p>\n\n\n\n After rounding to appropriate significant figures (based on 25.0 g, which has three significant figures), the result becomes:4.93 g of Al\\boxed{4.93\\ \\text{g of Al}}4.93 g of Al\u200b<\/p>\n\n\n\n This calculation reflects the proportional mass contribution of aluminum in the total compound. Aluminum makes up a fraction of the molar mass of Al\u2082(SO\u2084)\u2083, and the calculation isolates this contribution based on stoichiometric ratios.<\/p>\n\n\n\n How many grams of aluminum are present in 25.0 g of Al2(SO4)3? Molar mass( Al2(SO4)3) = 342.17 g\/mol The Correct Answer and Explanation is: Correct Answer:4.93 grams of aluminum are present in 25.0 g of Al\u2082(SO\u2084)\u2083. Explanation:To determine the mass of aluminum in a given amount of aluminum sulfate, begin by examining the formula: Al\u2082(SO\u2084)\u2083. 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4.93 grams of aluminum are present in 25.0 g of Al\u2082(SO\u2084)\u2083.<\/p>\n\n\n\n
\n\n\n\n
To determine the mass of aluminum in a given amount of aluminum sulfate, begin by examining the formula: Al\u2082(SO\u2084)\u2083<\/strong>. This compound contains 2 moles of aluminum (Al)<\/strong> per 1 mole of Al\u2082(SO\u2084)\u2083<\/strong>.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-584.jpeg\")
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