Chemical Calculations Problems Page 2
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1
2
3Chemistry 100
Clark College
5. Solid iron (III) oxide reacts with carbon monoxide gas to form iron and carbon dioxide gas.
a) Write a balanced reaction (including phase labels) for this reaction.
→ 2 Fe
Fe
O
+ 3 CO
+ 3 CO
2
3 (s)
(g)
(s)
2 (g)
b) Classify the reaction: is it single displacement or double displacement? Precipitation,
acid/base or redox?
It is probably double displacement. It is definitely redox and precipitation!
c) If 187 gram of iron (III) oxide is allowed to react with 105.8 grams of carbon monoxide,
how much iron (in grams) can be generated?
1 mol Fe
O
2 mol Fe
55.85 g Fe
Fe
O
: 187 g Fe
O
x
2
3
x
x
= 131 g Fe
2
3
2
3
159.7 g Fe
O
1 mol Fe
O
1 mol Fe
2
3
2
3
1 mol CO
2 mol Fe
55.85 g Fe
CO: 105.8 g CO x
x
x
= 141 g Fe
28.01 g CO
3 mol CO
1 mol Fe
Fe
O
is the limiting reagent, 131 g of Fe are produced.
2
3
6. Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at
high temperatures, as described by the following balanced equation.
(g) + 3 CuO (s) → N
2 NH
(g) + 3 Cu (s) + 3 H
O (g)
3
2
2
a) How many grams of N
are formed when 18.1 g of NH
are reacted with 90.4 g of CuO?
2
3
1 mol N
mol
28.01 g
2
18.1 g NH
x
x
x
= 14.9 g N
3
2
17.031 g
2 mol NH
1 mol
3
mol
1 mol N
28.013 g
! Limiting Reagent!
2
90.4 g CuO x
x
x
= 10.6 g N
2
79.545 g
3 mol CuO
mol
b) What starting material, if any is left over? How many grams of that material are left
over?
Since CuO is the limiting reagent, NH
is in excess. To determine this amount of excess,
3
determine how much extra N
would have been produced from the NH
, and convert that back
2
3
to g of NH
.
3
14.9 g - 10.6 g = 4.3 g excess N
2
1 mol
2 mol NH
17.031 g
3
4.3 g N
x
x
x
= 5.2 g of excess NH
2
3
28.02 g
1 mol N
1 mol
2
Chapter 8 Homework
Su07
Page 2 of 3
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