Question

How many moles of solid aluminum are needed to react with nitric acid in order to produce 5.0 moles of hydrogen gas?

Answer 1

A bit over `3*mol` aluminum are required...........

Explanation:

We need a stoichiometric equation that represents the oxidation of aluminum metal.....

`Al(s) + 3HNO_3(aq) rarr Al(NO_3)_3(aq) + 3/2H_2(g)uarr`

And thus each mole of metal generates `3/2` moles dihydrogen gas......And so if we want a `5*mol` quantity of `H_2` we take the quotient......

`(5*mol)/(3/2*mol)=10/3*mol` `"aluminum metal"`. Given the molar mass of aluminum is `27.0*g`, what quantity of metal are needed for the given reaction?