Question #0a43f

1 Answer
Jul 27, 2017

#"100 g O"_2#


For starters, it doesn't matter under what conditions for pressure and temperature you keep the oxygen gas, the mass required for this reaction to take place is always the same.

The volume of the gas that contains the needed number of grams of oxygen will depend on the conditions for pressure and temperature, but the mass will not.

So, you know that

#"C"_ 6"H"_ 12"O"_ (6(s)) + 6"O"_ (2(g)) -> 6"CO"_ (2(g)) + 6"H"_ 2"O"_ ((g))#

Notice that the reaction consumes #6# moles of oxygen gas for every #1# mole of glucose that takes part in the reaction.

Since glucose has a molar mass of #"180.156 g mol"^(-1)# and oxygen gas has a molar mass of #"32.0 g mol"^(-1)#, you can say that the reaction consumes

#6 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "192 g"#

of oxygen gas for every #"180.156 g"# of glucose that take part in the reaction.

This means that in order to completely consume #"90 g"# of glucose, you will need

#90 color(red)(cancel(color(black)("g C"_6"H"_12"O"_6))) * overbrace("192 g O"_2/(180.156color(red)(cancel(color(black)("g C"_6"H"_12"O"_6)))))^(color(blue)("the equivalent of the 1:6 mole ratio")) = color(darkgreen)(ul(color(black)("100 g O"_2)))#

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of glucose.