How many grams of HCl are consumed by the reaction of 2.50 moles of Magnesium, Mg?

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

3O2 + 4NH3 ---> 2N2 + 6H2O.

Explanation:

Let me give you (c) as an example: If you start out with 4.5 mol O2,  assuming you want to completely react all the oxygen you'd need to add how much ammonia? The ratio of moles of O2 versus moles of NH3 is

(moles NH3) / (moles O2) = 4/3, or in other words,

moles NH3 = (4/3) x moles O2.

Therefore you need to add (4/3) x (4.5 moles) = 6 moles NH3.  

The question says it want you to calculate the numbers of moles of all products and reactants for each case. So, that makes  

(c) 4.5 moles [O2] + 6 moles [NH3] > 3 moles [N2] + 9 moles [H2O].  

You need to do the same for (a) and (b)

2) The balanced net reaction is:  

Mg + 2HCl > MgCl2 + H2

According to this reaction, 2 moles of HCl are consumed for every mole of Mg added.

*tilts head* sounds more like a statement

imma come back to you hold on