# 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 "^oC when 0.498 g of aluminum react with an excess of HBr?

Nov 21, 2016

First you need a stoichiometric equation:

$A l \left(s\right) + 3 H B r \rightarrow A l B {r}_{3} \left(a q\right) + \frac{3}{2} {H}_{2} \left(g\right)$

Finally we get under a $\frac{1}{2} \cdot L$ volume of dihydrogen.

#### Explanation:

Each equiv of metal reduces $3$ equiv of acid, and $\frac{3}{2}$ equiv of dihydrogen gas are evolved.

$\text{Moles of aluminum}$ $=$ $\frac{0.498 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1} = 0.0185 \cdot m o l$.

Now, given the conditions, clearly the metal is the limiting reagent. We have the stoichiometry, so a molar quantity of $\frac{3}{2} \times 0.0185 \cdot m o l$ $\text{dihydrogen gas}$ result.

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

$V = \frac{n R T}{P} =$
(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.