We're asked to find the mass of #"O"_2# required to completely combust #2# #"g C"#.
To do this, we'll first write the balanced chemical equation for this reaction:
#"C"(s) + "O"_2(g) rarr "CO"_2(g)#
(note: if this were incomplete combustion, carbon monoxide would be produced instead)
Let's use the molar mass of carbon (#12.011# #"g/mol"#) to find the number of moles that burn:
#2cancel("g C")((1color(white)(l)"mol C")/(12.011cancel("g C"))) = color(red)(0.1665# #color(red)("mol C"#
Now, we can use the coefficients of the chemical equation to find the relative number of moles of #"O"_2# that burn:
#color(red)(0.1665)cancel(color(red)("mol C"))((1color(white)(l)"mol O"_2)/(1cancel("mol C"))) = color(green)(0.1665# #color(green)("mol O"_2#
Lastly, we'll use the molar mass of oxygen gas (#31.999# #"g/mol"#) to calculate the number of grams of oxygen:
#color(green)(0.1665)cancel(color(green)("mol O"_2))((15.999color(white)(l)"g O"_2)/(1cancel("mol O"_2))) = color(blue)(ul(3color(white)(l)"g O"_2#
which I suppose will leave at one significant figure (the more precise value is #color(blue)(2.66# #color(blue)("g O"_2#).
