Stoichiometry II :
Limiting Reagents
Or
When will we get there?
When are we going to stop?
Let’s consider the reaction of carbon and
oxygen to make carbon dioxide.
O2(g) 
C(s) +
12 g
1 mole
32 g
1 mole
CO2(g)
44 g
1 mole
So:
12 g C(s) + 32 g O2(g)  44 g CO2(g)
Let’s consider the reaction of carbon and
oxygen to make carbon monoxide.
2C(s) +
12 g
1 mole
O2(g) 
32 g
1 mole
2 CO (g)
28 g
1 mole
So:
24 g C(s) + 32 g O2(g)  56 g CO (g)
Next, let’s consider the reaction of hydrogen
and fluorine to make hydrogen fluoride.
H2(g) +
2 g
1 mole
F2(g)
38 g
1 mole
 2 HF(g)
20 g
1 mole
So:
2 g H2(g) + 38 g F2(g)  40 g HF(g)
Next, let’s consider the reaction of hydrogen
and oxygen to make water.
2 H2(g) + O2(g)  2 H2O(g)
2 g
1 mole
32 g
1 mole
18 g
1 mole
So:
4 g H2 + 32 g O2 
36 g H2O
Again, let’s consider the reaction of hydrogen
and oxygen to make water.
2 H2(g) + O2(g)  2 H2O(g)
2 g
1 mole
32 g
1 mole
18 g
1 mole
So:
4 g H2 + 34 g O2  36 g H2O + 2 g O2
2 H2(g) + O2(g)  2 H2O(g)
2 g
1 mole
32 g
1 mole
18 g
1 mole
4 g H2 + 34 g O2  ?
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4 g H2 1 mol H2 2 mol H2O 18 g H2O = 36 g H2O
2 g H2
2 mol H2 1 mol H2O
34 gO2 1 mol O2 2 mol H2O 18 g H2O = 38.25 g
32 g O2 1 mol O2 1 mol H2O H2O
So, in this reaction,
hydrogen is the limiting reagent.
When 4 grams of hydrogen have
been used the reaction stops
because all of the hydrogen is gone
32 grams of oxygen have been used,
so 2 grams of oxygen are left
4 g H2 + 34 g O2  36 g H2O + 2 g O2
4 g H2 + 32 g O2 + 2 g O2 36 g H2O + 2 g O2