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

1 Answer
Aug 2, 2017

Answer:

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?