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 $HBr$?

Md Ariful Islam

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Md Ariful Islam

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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 . We have the , 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.

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