Question #3fdb7

1 Answer
Mar 5, 2017


#"2960 g NH"_3#


Start by writing the balanced chemical equation that describes this reaction

#"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))#

The balanced chemical equation tells you that when the reaction consumes #3# moles of hydrogen gas and #1# mole of nitrogen gas, #2# moles of ammonia are produced.

The problem doesn't mention the mass of nitrogen gas you have at your disposal, so you can assume that this reactant is in excess, i.e. you have enough to ensure that all the moles of hydrogen gas react.

Now, you can convert the #3:2# mole ratio that exists between hydrogen gas and ammonia to a gram ratio by using the molar masses of the two gases.

You will have

#"3 moles H"_2/"2 moles NH"_3 = (3 color(red)(cancel(color(black)("moles H"_2))) * "2.016 g"/(1color(red)(cancel(color(black)("mole H"_2)))))/(2color(red)(cancel(color(black)("moles NH"_3))) * "17.03 g"/(1color(red)(cancel(color(black)("mole NH"_3))))) = (3 * "2.016 g")/(2 * "17.03 g")#


#"3 moles H"_2/"2 moles NH"_3 = "6.048 g H"_2/"34.06 g NH"_3#

You can now say that when #"525 g"# of hydrogen gas react with excess nitrogen gas, the reaction produces

#525 color(red)(cancel(color(black)("g H"_2))) * "34.06 g NH"_3/(6.048 color(red)(cancel(color(black)("g H"_2)))) = color(darkgreen)(ul(color(black)("2960 g NH"_3)))#

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