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?

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

#4"P"_text((s]) + color(red)(5)"O"_text(2(g]) -> 2"P"_2"O"_text(5(s])#

Notice that you have a #4:color(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 #5/4# time more moles of oxygen gas.

Use phosphorus' molar mass to determine how many moles you have in that #"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 #"23.0 g"# of oxygen for every #"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")#