If, in 2 M $\ce{HONO2}$, 2 mols $\ce{Al (s)}$ are oxidized to $\ce{Al(OH2)6^3+ (aq)}$, mols of reduction product hydrogen is
(a) two;
(b) three;
(c) four;
(d) five;
(e) six.
Regarding the above question, how would one solve it?
Would it be valid to simply say that I know that 6 electrons are being transferred because we have this skeleton equation (balanced for charge):
$$\ce{Al -> Al(OH2)6^3+ + 6 e-}$$
And given that 6 electrons are being transferred and that hydrogen is a reduction product, we must obtain hydrogen from some source in this system. The source would probably be hydronium ion, since hydronium ion is less stable than water. Plus abstracting protons from water to form hydrogen gas wouldn't make much sense in strongly acidic solution (we'd be forming hydroxide ions as well)! And knowing the following reaction:
$$\ce{2 H+ + 2 e- -> H2}$$
Can we just say since we know that 6 electrons are being moved, and that hydrogen gas is a product, we know that the immediately above reaction is occurring, and according to its stoichiometry, 2 moles of electrons yields 2 moles of hydrogen gas?