Questions dealing with:
Limiting Reactant
If 6 g of hydrogen and 70 g of oxygen react to form water:
(a) What is the limiting reactant?
(b)
How much water can be produced?
(c) How many grams of the excess reactant will be left
over?
First step is to write a balanced equation.
2
H
2
+ O
2
→
2
H
2
O
Second,
work a stoichiometric problem (which means you will have to use the
balanced equation to do the problem} with both of the reactants by themselves.
Let's do the problem using 6g of hydrogen first:
6 g H
2
X
1 mole H
2
X
2 moles H
2
O
*
X
18 g H
2
O
=
54 g of H
2
O
1
2 g H
2
2 moles H
2
1 mole H
2
O
*
The
2
and
2
in this step come from the coefficients in the balanced equation.
So, starting with 6 g of H
2
we can make 54 g of water.
Now let's so the problem using 70 g of
O
2
:
70g O
2
X
1mole O
2
X
2
moles of H
2
O
X
18g H
2
O
=
78.8 g H
2
O
1
32g O
2
1
mole O
2
1 mole H
2
O
So, starting with 70 g of O
2
we can make
78.8 g of H
2
O.
Since the hydrogen runs out after making 54 g of water, then 54 g is the amount of water
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 Spring '08
 TANIZAKI
 Chemistry, Mole, Sodium, H2O

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