83.9% is the weight percent of Al(C6H5)3 in the original 1.25 g sample.
First, look up the atomic weights of all elements involved.
Atomic weight of Aluminum = 26.981539
Atomic weight of Carbon = 12.0107
Atomic weight of Chlorine = 35.453
Atomic weight of Hydrogen = 1.00794
Now calculate the molar mass of Al(C6H5)3, and C6H6
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 15.999
= 258.293239 g/mol
Molar mass of C6H6 = 6 * 12.0107 + 6 * 1.00794
= 78.11184 g/mol
Determine how many moles of C6H6 was produced
0.951 g / 78.11184 g/mol = 0.012174851 mol
Since the balanced formula indicates that 3 moles of C6H6 is produced for each mole of Al(C6H5)3 used, divide by 3 to get the number of moles of Al(C6H5)3 that was present.
0.012174851 mol / 3 = 0.004058284 mol
Now multiply by the molar mass of Al(C6H5)3 to get the mass of Al(C6H5)3 originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, divide the mass of Al(C6H5)3 by the total mass of the original sample to get the weight percentage.
1.048227218 g / 1.25 g = 0.838582
Since all our measurements had 3 significant figures, round the result to 3 significant figures, giving 0.839 = 83.9%