Question

Question #a4ba2

Answer 1

`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."`