In the reaction: 2Al(s)+6HCl(aq) to 2Al^(3+)(aq) +6Cl^(-)(aq)+3H_(2)(g)

In the reaction: 2Al(s)+6HCl(aq) to 2Al^(3+)(aq) +6Cl^(-)(aq)+3H_(2)(

1.	In the reaction: 2Al(s)+6HCl(aq) to 2Al^(3+)(aq) +6Cl^(-)(aq)+3H_(2)(g)
Answer» 6L HCI (aq) is consumed for every 3L `H_2` (g) produced `33.6L H_2` (g) is produced REGARDLESS of temperature and pressure for every MOLE of AL that reacts `67.2L H_2` (g) at S.T.P. is produced for every mole of Al that reacts `11.2L H_2` (g) at S.T.P. is produced for every mole of HCI (aq-) consumed. Solution :`2Al(s) + underset("6 moles")(6HCl (aq)) to 2Al^(3+) (aq) + 6Cl^(-)(aq) + underset("3 moles" 3 xx 22.4 L "at S.T.P.")(3H_(2)(g))` `THEREFORE` 1 mole of HCl PRODUCES `=(3 xx 22.4)/6 = 11.2 L` of `H_(2)` at S.T.P.