A1.25-g sample contains some of the very reactive compound al(c6h5)3. on treating the compound with aqueous hcl, 0.951 g of c6h6 is obtained. al(c6h5)3(s) + 3hcl(aq) --> alcl3(aq) + 3c6h6(l) assuming that al(c6h5)3 was converted completely to products, what is the weight percent of al(c6h5)3 in original 1.25-g sample?

83.9% First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements. Atomic weight aluminum = 26.981539 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol Now determine how many moles of C6H6 was produced Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required. 0.012174851 mol / 3 = 0.004058284 mol Then multiply by the molar mass to get the number of grams that was originally present. 0.004058284 mol * 258.293239 g/mol = 1.048227218 g Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So 1.048227218 g / 1.25 g = 0.838581775 = 83.8581775% And of course, round to 3 significant digits, giving 83.9%

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