Stoichiometry Chemistry Worksheets With Answers Page 5
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11First write the reaction, then create and fill in your chart, and when you’re done filling in the
necessary blocks, it should look like this:
Molar mass
(23.95)
(44.01)
Balanced equation
2LiOH +
Li
CO
+
H
O
2
3
2
CO
2
No. moles
1000 g/23.95 g/mol = 41.8 moles
20.9
20.9
20.9
1.00 kg
Amount
20.9
44.01 = 920 g
1. Baking soda (NaHCO
) is often used as an antacid. It neutralizes excess hydrochloric acid
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secreted by the stomach in the reaction below:
NaHCO
+ HCl
NaCl
+ H
O
+ CO
3(s)
(aq)
(aq)
2
(l )
2(aq)
How many grams of NaHCO
would be needed to completely react with 10.0 g of HCl?
3
Explanation
According to the balanced equation above, 1 mole of NaHCO
reacts with 1 mole of HCl. First you
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need to calculate how many moles of HCl are in 10 grams of HCl. The formula weight of HCl is about
36 g/mol, so you set up the equation:
= 0.27 moles HCl.
Now, knowing that 1 mole of baking soda reacts with 1 mole of NaCl, we next need to figure out how
many grams of baking soda would react with 0.27 moles of baking soda: 0.27 moles
134 g/mol
baking soda = 37.2 grams/mole NaHCO
3
(85.32)
(36.46)
Molar mass
NaHCO
+
NaCl +
H
O +
CO
Balanced equation
3
2
2
HCl
No. moles
0.274
10.0/36.46 = 0.274
10.0 g
Amount
0.274
85.32 = 23.4 g
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Limiting Reagents
Have you ever noticed that hot dogs are sold in packages of 10, while the buns come in packages of
eight? In this scenario, the buns are the limiting reactant in the sense that they limit the hot dog
preparation to eight. The limiting reactant or reagent is the one that is consumed first in the
chemical reaction, and its consumption halts the progress of the forward reaction.
When answering questions about limiting reagents on the exam, your first step should always be to
convert all the masses you were given into moles. You should set up your table as you did before,
only now you’ll have two amounts and thus two numbers of moles to get you started.
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