# Suppose a sample of air contains 23.0% oxygen by mass. How many grams of air are required to complete the combustion of 93.0 g of P to diphosphorus pentoxide?

Nov 11, 2015

$\text{522 g}$

#### Explanation:

Your starting point here will be the balanced chemical equation for this combustion reaction

$4 {\text{P"_text((s]) + color(red)(5)"O"_text(2(g]) -> 2"P"_2"O}}_{\textrm{5 \left(s\right]}}$

Notice that you have a $4 : \textcolor{red}{5}$ mole ratio between phosphorus and oxygen. This means that, regardless of how many moles of phosphorus you have, the reaction will always need $\frac{5}{4}$ time more moles of oxygen gas.

Use phosphorus' molar mass to determine how many moles you have in that $\text{93.0-g}$ sample

93.0color(red)(cancel(color(black)("g"))) * " 1mole P"/(30.974color(red)(cancel(color(black)("g")))) = "3.0025 moles P"

Use the aforementioned mole ratio to determine how many moles of oxygen you would need for many moles of phosphorus to completely take part in the reaction

3.0025color(red)(cancel(color(black)("moles P"))) * (color(red)(5)" moles O"_2)/(4color(red)(cancel(color(black)("moles P")))) = "3.753 moles O"_2

Now use oxygen's molar mass to determine how many grams of oxygen would contain this many moles

3.753color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "120.1 g O"_2

Now, you know that your air sample contains 23% oxygen by mass, which means that you get $\text{23.0 g}$ of oxygen for every $\text{100 g}$ of air.

This means that you will need

120.1color(red)(cancel(color(black)("g O"_2))) * "100.0 g air"/(23.0color(red)(cancel(color(black)("g O"_2)))) = "522.17 g air"

Rounded to three sig figs, the answer will be

m_"air" = color(green)("522 g")