Question #a4ba2

1 Answer
Oct 17, 2017

2.78 moles of "H"_2

Explanation:

2 "Al" + 6 "HCl" = 2 "AlCl"_3 + 3 "H"_2

Here, you need 2 moles of Al and 6 moles of HCl to form 2 moles of AlCl3 and 3 moles of H2.

50cancel(g" Al")xx( 2 " mol Al")/(27 cancel(g " Al"))=3.7 " mol Al"

3.7 cancel(" mol Al") xx(6 "mol HCl")/(2cancel(" mol Al"))=11.1" mol HCl reqd."

11.1 cancel(" mol HCl") xx(3 "mol H"_2)/(2cancel(" mol HCl"))=16.65" mol H"_2" produced."