Question #e44f6

Write down the balanced chemical equation for this reaction and take a look at the mole ratio that exists between the two reactants

#"N"_ (2(g)) + color(red)(2)"O"_ (2(g)) -> color(blue)(2)"NO"_(2(g))#

As you can see, the reaction consumes #color(red)(2)# moles of oxygen gas for every mole of nitrogen gas that takes par in the reaction.

The problem provides you with the number of moles of nitrogen gas available for the reaction. Use the aforementioned mole ratio to determine how many moles of oxygen gas would be needed in order to make sure that all the moles of nitrogen as raect

#1.25color(red)(cancel(color(black)("moles N"_2))) * (color(red)(2)color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole N"_2)))) = "2.50 moles O"_2#

At this point, you can say that if the #"50.0-g"# sample of oxygen given to you contains fewer than #2.50# moles, then oxygen gas will be limiting reagent.

If that sample contains more than #2.50# moles, then oxygen will be in excess, i.e. nitrogen gas will be the limiting reagent.

Use oxygen gas' molar mass to determine how many moles of oxygen gas you have available

#50.0 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "1.563 moles O"_2#

As you can see, you don't have enough oxygen gas to ensure that all the moles of nitrogen gas take part in the reaction, so

Oxygen gas, #"O"_2#, will be your limiting reagent

Now, the reaction will completely consume the oxygen gas and leave you with excess nitrogen gas.

Notice that you have a #color(red)(2):color(blue)(2)# mole ratio between oxygen gas and nitrogen dioxide. This tells you that the reaction produces

#1.563color(red)(cancel(color(black)("moles O"_2))) * (color(blue)(2)color(white)(a)"moles NO"_2)/(color(red)(2)color(red)(cancel(color(black)("moles O"_2)))) = "1.563 moles NO"_2#

To convert this to grams, use nitrogen dioxide's molar mass

#1.563color(red)(cancel(color(black)("moles NO"_2))) * "46.006 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)"71.9 g"color(white)(a/a)|)))#

The answer is rounded to three sig figs.