A 1.25-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?

I got 83.8%. close enough. just make sure not to round your values until the very end (keep them in your calc).

Is it 84.8%?

To find the weight percent of Al(C6H5)3 in the original 1.25-g sample, we need to determine the mass of Al(C6H5)3 in the sample.

First, let's calculate the mass of C6H6 obtained from the reaction. We know that 0.951 g of C6H6 was obtained. Since the molar ratio between Al(C6H5)3 and C6H6 is 1:3, we can calculate the mass of Al(C6H5)3 using the following equation:

(0.951 g C6H6) x (1 mol Al(C6H5)3 / 3 mol C6H6) x (306.34 g Al(C6H5)3 / 1 mol Al(C6H5)3)

The molar mass of C6H6 is 78.11 g/mol, and the molar mass of Al(C6H5)3 is 306.34 g/mol.

Now we can calculate the mass of Al(C6H5)3:

(0.951 g C6H6) x (1 mol Al(C6H5)3 / 3 mol C6H6) x (306.34 g Al(C6H5)3 / 1 mol Al(C6H5)3) = 0.101 g Al(C6H5)3

Next, we can calculate the weight percent:

(weight of Al(C6H5)3 / weight of original sample) x 100

(0.101 g / 1.25 g) x 100 = 8.08%

Therefore, the weight percent of Al(C6H5)3 in the original 1.25-g sample is 8.08%.

This is a stoichiometry problem. Convert, using the coefficients in the balanced equation, 0.951 g C6H6 product to grams Al(C6H5)3. Then %Al(C6H5)3 = [g Al(C6H5)3/1.25)]*100 = ?

I don't understand how you got to 83.8%??