Aluminum reacts with hydrochloric acid according to the equation. #2Al_((g)) + 6HCl_((aq)) -> 2AlCl_(3(aq)) +3H_(2(g))#. Calculate the volume of 1.50mol/L hydrochloric acid that is required for 5.40g of Aluminum to react completey?

1 Answer
Dave
Apr 27, 2015

I think the aluminum in your equation should be solid (s).

mass of aluminum to moles of aluminum to moles of HCl (using balanced equation) to volume of HCl

5.40g Al (1mole Al/26.98g) = 0.200 moles Al

0.200 moles Al (6moles HCl / 2moles Al) = 0.600 moles HCl

Vol HCl = 0.600 moles HCl / 1.5M HCl = 0.400 Liters HCl

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