Answer to Question #133988 in General Chemistry for smntc

Question

In a bowl throw 2.7 grams of aluminum in powder form and ignite it in air. the product obtained by combustion is dissolved with dilute hydrochloric acid. calculate how many grams of salt will be obtained from this case.

Answer 1

4Al+3O₂=2Al₂₃O₃

Al₂O₃+6HCl=2AlCl₃+3H₂O

according to the above reactions:

4Al=> 2Al₂O₃

2Al₂O₃=> 4AlCl₃ that is,

1Al=> 1AlCl₃

1. According to the above reaction, 27 g (1 mol) of Al produces 133.5 g (1 mol) of AlCl₃. Find how much AlCl₃ is formed from 2.7 g (0.1 mol) of Al:

27 gr Al — 133.5 gr AlCl₃

2.7 gr Al — x gr AlCl₃

x= 2.7*133.5/27= 13.35 gr

ANSWER: 13.35 gr AlCl₃