# Question 9dda7

Jul 9, 2017

Nitric oxide is the limiting reagent.

#### Explanation:

Start by using the ideal gas law equation to find the number of moles of oxygen gas present in the sample. I take it that you're already familiar with the ideal gas law equation, so I won't go into too much detail about it here.

You know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

so rearrange to solve for $n$

$n = \frac{P V}{R T}$

Plug in your values to find--do not forget to convert the temperature to Kelvin and the pressure to atm before using the values you have for the temperature and pressure of the gas!

$n = \left(\frac{631}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 85.3 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 39.0)color(red)(cancel(color(black)("K}}}}\right)$

$n = {\text{2.7635 moles O}}_{2}$

Now, to convert the mass of nitric oxide to moles, you must use the compound's molar mass.

157 color(red)(cancel(color(black)("g"))) * "1 mole NO"/(30.006 color(red)(cancel(color(black)("g")))) = "5.232 moles NO"

You know from the balanced chemical equation that the two reactants react in a $2 : 1$ mole ratio.

$2 {\text{NO"_ ((g)) + "O"_ (2(g)) -> 2"NO}}_{2 \left(g\right)}$

This tells you that in order to take part in the reaction, every $1$ mole of oxygen gas requires $2$ moles of nitric oxide.

In your case, $2.7635$ moles of oxygen gas would require

2.7635 color(red)(cancel(color(black)("moles O"_2))) * "2 moles NO"/(1color(red)(cancel(color(black)("mole O"_2)))) = "5.527 moles NO"

Since you have fewer moles of nitric oxide than you need to ensure that all the moles of oxygen gas can take part in the reaction

overbrace("5.527 moles NO")^(color(blue)("what you need")) " "> " " overbrace("5.232 moles NO")^(color(blue)("what you have"))

you can say that nitric oxide will act as a limiting reagent, i.e. it will be completely consumed before all the moles of oxygen gas will get the chance to react.

This is, of course, equivalent to saying that oxygen gas is in excess.

You can thus say that the reaction will consume $5.232$ moles of nitric oxide and only

5.232 color(red)(cancel(color(black)("moles NO"))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles NO")))) = "2.616 moles O"_2#

out of the $2.7635$ moles of oxygen gas available.