Aluminum reacts with aqueous #HBr# to produce hydrogen gas and aluminum bromide. How many mL of hydrogen gas are produced at 1.03 atm and 20.00 #"^o#C when 0.498 g of aluminum react with an excess of #HBr#?

1 Answer
Nov 21, 2016

Answer:

First you need a stoichiometric equation:

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

Finally we get under a #1/2*L# volume of dihydrogen.

Explanation:

Each equiv of metal reduces #3# equiv of acid, and #3/2# equiv of dihydrogen gas are evolved.

#"Moles of aluminum"# #=# #(0.498*g)/(26.98*g*mol^-1)=0.0185*mol#.

Now, given the conditions, clearly the metal is the limiting reagent. We have the stoichiometry, so a molar quantity of #3/2xx0.0185*mol# #"dihydrogen gas"# result.

And now it is an Ideal Gas calculation (mind you if we collect the gas over water, we should include the #"saturated vapour pressure"#, to account for the water vapour collected with the dihydrogen. Since there are no details in the initial conditions, I can ignore these data.)

#V=(nRT)/P=#
#(0.0185*molxx0.0821*L*atm*K^-1*mol^-1xx293.15*K)/(1.03*atm)=0.431*L=??*mL#.

This is experiment that could be performed in a high school lab with litre graduated cylinders, with which to collect the gas.