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

Question

Caleb Adams · Accepted Answer

#2.5mol#

First, create a balanced chemical equation to describe how the metal and acid will react:

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

This is an illustration of the mole ratio at which the reactants mix and create the products.

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#

Ellie Andrews · Answer

#2.5# #mol# dihydrogen gas are formed.

#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.

Luna Chen · Answer

undefined The balanced chemical equation for the reaction between aluminum (Al) and hydrogen bromide (HBr) is:

2Al + 6HBr -> 2AlBr3 + 3H2

From the balanced equation, it can be inferred that 3 moles of H2 are produced for every 2 moles of aluminum reacting.

Given that 4.22 moles of aluminum react, we use the stoichiometry of the reaction to determine the moles of H2 produced:

(4.22 moles Al) * (3 moles H2 / 2 moles Al) = 6.33 moles H2

Therefore, 6.33 moles of H2 are formed.