Academic Subjects
➤
Chemistry
➤
General Chemistry
➤
Chemical Reactions
➤
Reacting 120 Moles H
2
With 50 Moles N
2
Ammonia is formed from the reaction of hydrogen and nitrogen.
3 H
2 (g)
+ N
2 (g)
⇒
2 NH
3 (g)
How many grams of which reagent remains theoretically after
120 moles of hydrogen are reacted with 50 moles of nitrogen?
We already have our balanced equation, so let's find the limiting reagent
(the one that we will run out of first)
ammonium
120
moles H
2
(2 moles NH
3
)
/
(3
moles H
2
)
=
80 moles NH
3
limiting reagents
50
moles N
2
(2 moles NH
3
)
/
(1
moles N
2
)
=
100 moles NH
3
ammonia
Based on these calculations, H
2
is the limiting reagent
Let's find out how much N
2
we need to make 80 moles NH
3
80
moles NH
3
(1 moles N
2
)
/
(2
moles NH
3
)
= 40 moles N
2
In other words, to react all of the H
2
we only need 40 moles of N
2
This means 10 moles of N
2
will be left over after the reaction is completed
Let's turn these 10 moles of N
2
into grams
10
moles N
2
(28.00 g N
2
)
/
(1
mole N
2
)
= 280 g N
2
Once this reaction is complete, there will be 280 g N
2
left over