Question

What mass of HCI is needed to

generate 45.2 g of AICI3?
2AI + 6HCI → 2AICI3 + 3H2
AICI3: 133.33 g/mol
HCI: 36.46 g/mol
[?] g AlCl3

Answer 1

you must use the process known as dimensional analysis. start with what the problem gives you, which in this case is g of AlCl3!
(45.2 g AlCl3) * (1 mol AlCl3/133.33g AlCl3) * (6mol HCl/2mol AlCl3) * (36.46gHCl/1molHCl)
Now multiply across. the correct answer should cancel all units except g HCl, since that is what the problem asks for
34.9 g HCl is required!