Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a

dark-brown gas:
2NO(g) + O2(g) -> 2NO2(g)
In one experiment 0.886 moles of NO is mixed with 0.503 moles of O2.
Calculate which of the two reactants is the limiting reagent.

Please show set up of problem and show work.

To determine which reactant is the limiting reagent, we need to compare the moles of each reactant and see which one is more limited in the reaction.

Given:
Moles of NO = 0.886 moles
Moles of O2 = 0.503 moles

To calculate the limiting reagent, we will use the stoichiometry of the balanced chemical equation.

The balanced equation is:
2NO(g) + O2(g) -> 2NO2(g)

From the balanced equation, we can see that the ratio of NO to O2 consumed is 2:1.

1) Calculate the number of moles of NO required to react with O2:
Moles of O2 * (2 moles of NO / 1 mole of O2) = 0.503 moles * (2 moles of NO / 1 mole of O2) = 1.006 moles of NO required

2) Compare the calculated moles of NO required with the actual moles of NO given:
Moles of NO required = 1.006 moles
Moles of NO given = 0.886 moles

Since the actual moles of NO given (0.886 moles) are less than the moles of NO required (1.006 moles), it means that NO is the limiting reagent.

Therefore, in this reaction, NO is the limiting reagent, and O2 is the excess reagent.