When 4.22 moles of #Al# reacts with 5.0 moles of #HBr#, how many moles of #H_2# are formed?

2 Answers
Trevor Ryan.
Mar 26, 2016

#2.5mol#

Explanation:

First write a balanced chemical equation for the reaction of the metal with the acid :

#2Al+6HBr->2AlBr_3+3H_2#.

This represents the mole ratio in which the reactants combine and in which the products are formed.

Since #2# moles #Al# requires #6# moles #HBr#, it implies that, given #4.22# moles #Al# and #5.0# moles #HBr#, then #HBr# is in excess and #Al# is the limiting reagent.

Hence all #5.0 mol HBr# will react with #5.0/3=1.667mol Al# to form #1.667mol AlBr_3# and #5.0/2=2.5 mol H_2#

anor277
Mar 26, 2016

#2.5# #mol# dihydrogen gas are formed.

Explanation:

#Al(s) + 3HBr(aq) rarr AlBr_3(aq) + 3/2H_2(g)uarr#

Clearly, hydrobromic acid is in deficiency (complete rxn requires #>12*mol# #HBr#). From the equation, #1/2# #mol# dihydrogen gas are evolved per mole of #HBr#. Since #5# #mol# #HBr# are used, half of this quantity of #H_2# evolve.

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