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

Jan 14, 2016

$N O \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow N {O}_{2} \left(g\right)$

Explanation:

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, $N O$, combines with stoichiometic oxygen, to give stoichiometric nitric oxide, $N {O}_{2}$. 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.