If you began with 9.23 moles of hydrochloric acid and excess amounts of aluminum, how many grams of aluminum chloride would be formed?

This is a stoichometry problem, in which you use the relationship between reactants and/or products in a chemical equation to calculate the quantity of one substance base on the quantity of any other substance in the reaction. These relationships are given by the coefficients in the balanced equation, so it's necessary to begin by writing and balancing the equation. The equation for this reaction is:

`2 Al + 6 HCl -gt 2 AlCl_3 + 3...

This is a stoichometry problem, in which you use the relationship between reactants and/or products in a chemical equation to calculate the quantity of one substance base on the quantity of any other substance in the reaction. These relationships are given by the coefficients in the balanced equation, so it's necessary to begin by writing and balancing the equation. The equation for this reaction is:

`2 Al + 6 HCl -gt 2 AlCl_3 + 3 H_2`

The mole ratio of AlCl3 to HCl is 2:6, so the moles of AlCl3 formed is:

9.23 moles HCl x (2 AlCl3)/(6 HCl) =  3.08 moles AlCl3

Multiplying moles of AlCl3 by its molar mass gives grams:

3.08 moles AlCl3 x 133 grams/1 mole = 410. grams AlCl3

You can solve problems like this more quickly be multiplying the given quantity by a series of conversion factors, without having to calculate the intermediate step(s):

9.23 moles HCl x (2 AlCl3)/(6 HCl) x 133g/1mol = 410. grams AlCl3