Question

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`?

Answer 1

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.