Question #793e1

You know that ammonia is in excess, so right from the start, you can say that the number of moles of nitric oxide, #"NO"#, produced by the reaction will depend exclusively on the number of moles of oxygen gas consumed by the reaction.

#4"NH"_ (3(g)) + color(blue)(5)"O" _ (2(g)) -> color(purple)(4)"NO"_ ((g)) + 6"H"_ 2"O"_ ((g))#

Notice that the reaction produces #color(purple)(4)# moles of nitric oxide for every #color(blue)(5)# moles of oxygen gas that take part in the reaction.

Now, the problems gives you grams of oxygen gas and asks for grams of nitric oxide, so one approach to use here would be to convert the mole ratio to a gram ratio.

To do that, use the molar masses of the two chemical species

#(color(blue)(5)color(white)(.)"moles O"_2)/(color(purple)(4)color(white)(.)"moles NO") = (color(blue)(5) color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))))/(color(purple)(4)color(red)(cancel(color(black)("moles NO"))) * "30.0 g"/(1color(red)(cancel(color(black)("mole NO"))))) = "160 g O"_2/"120 g NO" = = "4 g O"_2/"3 g NO"#

This means that for every #"4 g"# of oxygen gas that take part in the reaction, you get #"3 g"# of nitric oxide.

You can thus say that your reaction will produce

#32.2 color(red)(cancel(color(black)("g O"_2))) * "3 g NO"/(4color(red)(cancel(color(black)("g O"_2)))) = color(darkgreen)(ul(color(black)("24.2 g NO")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of oxygen gas.