Using the equation #"N"_2 + 3"H"_2 - 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced?

Question

Using the equation #"N"_2 + 3"H"_2 -> 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced?

Ellie Andrews · Accepted Answer

#"34 g NH"_3#

The key to any stoichiometry problem is the mole ratio that exists in the balanced chemical equation between the two chemical species that are of interest.

In this case, you have

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

The balanced chemical equation tells you that the reaction consumes #3# moles of hydrogen gas, #"H"_2#, and produces #color(red)(2)# moles of ammonia, #"NH"_3#, for every mole of nitrogen gas, #"N"_2#, that takes part in the reaction.

Since the problem doesn't mention the amount of hydrogen gas available for the reaction, you can assume that it's in excess, which means that the reaction will consume all the grams of nitrogen gas present.

Now, you can convert the mole ratio to a gram ratio by using the molar masses of the two species

#M_("M N"_2) ~~ "28 g mol"^(-1)#

#M_("M NH"_3) ~~ "17 g mol"^(-1)#

This means that the #1:color(red0(2)# mole ratio that exists between nitrogen gas and ammonia can be written as

#(1 color(gray)(cancel(color(black)("mole"))) * "28 g" color(gray)(cancel(color(black)("mol"^(-1)))))/(color(red)(2) color(gray)(cancel(color(black)("moles"))) * "17 g" color(gray)(cancel(color(black)("mol"^(-1))))) = (28 color(gray)(cancel(color(black)("g"))))/(34color(gray)(cancel(color(black)("g")))) = 14/17 -># gram ratio

So, if the reaction consumes #"28 g"# of nitrogen gas, it follows that it will produce

#28 color(red)(cancel(color(black)("g N"_2))) * "17 g NH"_3/(14color(red)(cancel(color(black)("g N"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("34 g NH"_3)color(white)(a/a)|)))#

In other words, #1# mole of nitrogen gas will produce #color(red)(2)# moles of ammonia.

The answer is rounded to two sig figs.

Jose Ayala · Answer

undefined To find the grams of NH3 produced, use the stoichiometry of the reaction:

1 mol N2 reacts to produce 2 mol NH3.

First, convert grams of N2 to moles:
28g N2 * (1 mol N2 / 28 g N2) = 1 mol N2

Then, use the stoichiometry to find moles of NH3 produced:
1 mol N2 * (2 mol NH3 / 1 mol N2) = 2 mol NH3

Finally, convert moles of NH3 to grams:
2 mol NH3 * (17 g NH3 / 1 mol NH3) = 34 g NH3

Therefore, 34 grams of NH3 will be produced.

Using the equation #"N"_2 + 3"H"_2 -&gt; 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced?

Using the equation #"N"_2 + 3"H"_2 -> 2"NH"_3#, if 28g #"N"_2# react, how many grams of #"NH"_3# will be produced?