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Chapter 3: Formulas, Equations, and Moles
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Self Quiz 2
Self Quiz 2
This activity contains 20 questions.
In order to balance the equation C
2
H
6
+ O
2
---> H
2
O + CO
2
you should
add H
2
to the products to balance H.
change the subscript of O in water to 2 to help balance the O.
add O
2
to the product side to help balance the O in the equation.
change the coefficients.
add C(
s
) to the product side to balance the C.
Which equation would be a good way of using chemical symbols to represent this description of this chemical reaction? A formula unit of phosphoric acid reacts with 3 formula units of sodium hydroxide to yield a formula unit of sodium phosphate and 3 molecules of water.
H
3
PO
4
+ NaOH ---> Na
3
PO
4
+ H
2
O
H
3
PO
4
+ 3NaOH ---> Na
3
PO
4
+ 3H
2
O
H
3
PO
4
---> 3NaOH + Na
3
PO
4
+ 3H
2
O
Na
3
PO
4
+ 3H
2
O ---> H
3
PO
4
+ 3NaOH
98 g of H
3
PO
4
+ 120 g of NaOH ---> 164 g of Na
3
PO
4
+ 54 g of H
2
O
Complete this statement: _____________ are in 10.0 mole of C
10
H
8
.
10.0 moles of C
4.82 x 10
25
atoms of H
6.022 x 10
24
atoms of C
8.00 moles of H
4.818 x 10
24
atoms of H
Which statement is the correct interpretation for the equation showing the rusting of iron: 4Fe + 3O
2
---> 2Fe
2
O
3
?
4 g of iron + 3 g of oxygen will produce 2 g of iron (III) oxide.
223.4 g of iron + 48.00 g of oxygen will produce 271.4 g of iron (III) oxide.
55.85 g of iron + 32.00 g of oxygen will produce 159.7 g of iron (III) oxide.
223.4 g of iron + 96.00 g of oxygen will produce 319.4 g of iron (III) oxide.
4 g of iron + 6 g of oxygen will produce 10 g of iron (III) oxide.
Which statement is
not
true?
10.0 moles of water contain 6.022 x 10
24
molecules of water.
10.0 moles of water is equivalent to 18.02 g of water.
10.0 moles of water contain the same number of O atoms as 5.00 moles of CO
2
.
10.0 moles of water contain the same number of molecules as 10.0 moles of H
2
.
10.0 moles of water contain the same number of H atoms as 2.00 moles of butane (C
4
H
10
).
According to the equation I
2
+ 3Cl
2
---> 2ICl
3
, how many grams of ICl
3
are produced if 1.00 g of Cl
2
reacts with an excess of I
2
?
9.89 g
3.29 g
4.93 g
4.39 g
2.19 g
The reaction C
7
H
8
+ 3HNO
3
---> C
7
H
5
N
3
O
6
+ 3H
2
O can be used to make TNT. How many grams of HNO
3
are required to react with 10.0 g of C
7
H
8
?
2.28 g
6.84 g
10.1 g
20.5 g
30.0 g
Suppose 10.0 g of CH
4
are used in a reaction that can theoretically produce 95.8 g of CCl
4
, a former dry-cleaning solvent. What is the percent yield of CCl
4
if the reaction produces only 80.0 g of CCl
4
?
80.0 %
15.8 %
83.5 %
10.4 %
16.5 %
What is the percent yield of CaO in the reaction CaCO
3
---> CaO + CO
2
if 5.33 g of CaO are obtained when 10.0 g of CaCO
3
are used?
5.60 %
53.3 %
64.7 %
5.33 %
95.2 %
If 4.0 moles of Li and 2.0 moles of O
2
are used in the reaction 4Li + O
2
---> 2Li
2
O, then the limiting reactant is _________ and the theoretical yield of Li
2
O is ____________ g.
oxygen, 1.2 x 10
2
lithium, 6. 0 x 10
oxygen, 6.0 x 10
lithium, 3.0 x 10
oxygen, 3.0 x 10
How many moles of HCl are in 20.0 mL of 0.200 M HCl solution?
0.0100 mole
10.0 moles
4.00 moles
0.00400 moles
0.0400 moles
When a 100 mL sample of 1.0 M NaCl is diluted to form 1.0 L of 0.10 M NaCl which statement is
not
true?
The number of moles of NaCl in the concentrated solution is equal to the number of moles of NaCl in the dilute solution.
The equation M
i
x V
i
= M
f
x V
f
could be used in this procedure to determine the volume of the dilute solution.
As the volume of the solution increases by a factor of 10, the concentration decreases by a factor of 10.
When diluting the 1.0 M NaCl, 900 mL of water are added.
No solute is removed to form the dilute solution.
Which of these gives a dilute solution of KNO
3
having a concentration of 0.150 M?
0.100 L of a 3.00 M KNO
3
solution is diluted to 200.0 mL.
120.0 mL of a 0.400 M KNO
3
solution is diluted to 3.20 L.
3.00 mL of 1.00 M KNO
3
solution is diluted to 200.0 mL.
6.07 g of KNO
3
is added to enough water to make 4.00 L of solution.
1.52 g of KNO
3
is added to enough water to make 100.0 mL of solution.
Citric acid, H
3
C
6
H
5
O
7
, can be neutralized by reaction with NaOH, according to the equation H
3
C
6
H
5
O
7
(
aq
) + 3NaOH(
aq
) ---> 3H
2
O(
l
) + Na
3
C
6
H
5
O
7
(
aq
). The number of milliliters of 0.150 M NaOH required to neutralize 200.0 mL of 0.100 M citric acid is
4.00 x 10
2
mL.
44.4 mL.
40.0 mL.
133 mL.
9.00 x 10
2
mL.
The number of grams of Al that will react with 75.0 mL of 0.875 M HCl, according to the equation 2Al + 6HCl ---> 3H
2
+ 2AlCl
3
is
1.77 g.
0.771 g.
0.590 g.
5.31 g.
5.90 x 10
2
.
A compound that is made of 33.3% carbon and 66.7% oxygen
has the empirical formula CO
2
.
contains 0.333 mole C and 0.667 mole O in a 1-mole sample.
contains 0.333 g C and 0.667 g O in a 1-g sample.
contain 33 atoms of C and 67 atoms of O in 1 molecule.
contains 4.00 g C and 10.7 g O in a 14.7-g sample.
The percent composition of the compound with the formula C
2
H
4
O
2
is
40.0% C, 6.73% H, 53.3% O.
25.0% C, 50.0% H, 25.0% O.
41.4% C, 3.45% H, 55.1% O.
20.0% C, 1.68% H, 26.6% O.
37.5% C, 12.5% H, 50.0% O.
Which statement concerning the molecular formula of a substance is
not
true ?
It gives a whole-number ratio of atoms in the molecule.
It can always be determined from the percentage composition of the compound.
It can always be determined from the percent composition of the compound and the molar mass.
It gives very little, if any, information about how the atoms in the molecule are connected.
It gives the actual number and type of atoms in the molecule.
On combustion analysis, a 100.0 g-sample of a compound containing only C and H produced 313 g of CO
2
(44.01 g/mole) and 128 g of H
2
O (18.02 g/mole). The empirical formula for this compound is
C
5
H
2
.
C
7
H
7
.
CH.
C
2
H.
CH
2
.
A mass spectrum of a compound obtained from a mass spectrometer shows 5 peaks at 100 amu, 122 amu, 137 amu, 194 amu, and 212 amu. According to this data, the most probable molecular mass for this compound is
100 amu.
153 amu.
212 amu.
765 amu.
137 amu.
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