How many moles of nitrogen dioxide are produced from 4.754 mol NO? Given the equation: NO(g) + O2(g) --> NO2(g).

1 Answer
Jan 14, 2016

#NO(g) + 1/2O_2(g) rarr NO_2(g)#


The reaction above is stoichiometric. What does this mean? It means that for every product particle, there is a corresponding reactant particle; indeed there must be because masses are conserved in every chemical reaction.

Now of course I cannot have 1/2 an oxygen molecule, but I can certainly have 16 g of oxygen, which represents 1/2 a mole (i.e. Avogadro's number) of oxygen, #O_2#, molecules.

Given the stoichiometry, 1 mole of nitrous oxide, #NO#, combines with stoichiometic oxygen, to give stoichiometric nitric oxide, #NO_2#. So there is a 1:1 equivalence between nitrous oxide and its oxidation product.

You have started with 4.754 moles of nitrous oxide. Given the 1:1 stoichiometry, with how many moles of nitric oxide will you finish given sufficient oxygen gas? I think the answer is pretty clear.