The amount of aluminum in an ore sample was determined by dissolving the ore in excess hydrochloric acid acccording to the following reaction . The excess was then neutralized with sodium hydroxide.
3Al + 6HCL -> 2 AlCl3 + 3H2.
A 2.6 f ore sample was dissolved in 175 ml of 1.2 M hCl. This solution was diluted to a total volume of 500 mL. A 25 mL allquot of the diluted solution required 20.8 mL of 0.248 M NaOH for the excess HCl to be neutralized. What is the mass percent of aluminum in the ore sample?