Ethyne is completely combusted as per the following equation. If #2.40 xx 10^(-1) # moles of oxygen react, how many moles of carbon dioxide form? (Answer in mol to 3 s.d. in proper scientific notation.)

Look at the balanced chemical equation that describes this reaction

#"C"_ 2"H"_ (4(g)) + color(blue)(3)"O"_ (2(g)) -> color(purple)(2)"CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))#

Notice that when the reaction consumes #color(blue)(3)# moles of oxygen gas, #color(purple)(2)# moles of carbon dioxide are being produced.

This tells you that regardless of how many moles of oxygen gas react, the reaction will always produce carbon dioxide in a #color(blue)(3):color(purple)(2)# mole ratio.

In your case, you know that the reaction consumes

#n_( "O"_ 2) = 2.40 * 10^(-1)# #"moles O"_2#

You can now use the aforementioned mole ratio to determine how many moles of carbon dioxide will be produced

#2.40 * 10^(-1) color(red)(cancel(color(black)("moles O"_2))) * (color(purple)(2)color(white)(.)"moles CO"_2)/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2)))) = color(darkgeen)(ul(color(black)(1.60 * 10^(-1)color(white)(.)"moles CO"_2)))#

The answer is rounded to three sig figs and in proper scientific notation.

So remember, the stoichiometric coefficients that each compound has in the balanced chemical equation give you the mole ratio that exists when said compound are consumed or produced by the reaction.