Asked by simon
Al(s) + HCl(aq) → AlCl3(aq) + H2(g) According to the equation above, how many grams of aluminum are needed to completely react with 2.19 mol of hydrochloric acid?
Answers
Answered by
DrBob222
I can give an answer "according to the equation above" but it won't mean anything because the equation above is not balanced. Let me balance it and answer correctly.
2Al + 6 HCl ==> 2AlCl3 + 3H2
2.19 moles HCl will require 2.19 x (2 moles Al/6 moles HCl) = 2.19*2/6 = 0.73 moles Al. Convert that to grams Al; i.e.., grams Al = mols Al x atomic mass Al = ?
2Al + 6 HCl ==> 2AlCl3 + 3H2
2.19 moles HCl will require 2.19 x (2 moles Al/6 moles HCl) = 2.19*2/6 = 0.73 moles Al. Convert that to grams Al; i.e.., grams Al = mols Al x atomic mass Al = ?
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