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Toxins Unit – Investigation 5
Lesson 4:
Stoich Quiz Review 2
Admit Slip
How many grams of water will be formed from 290
g of C4H10 ?
2 C4H10
+ 13 O2  8 CO2
290 g C4H10
10 H2O
= 450 g H2O
÷ (58 g/mol)
5 mol C4H10
+
x (18 g/mol)
*
10molH 2O
2molC 4 H10
= 25 mol H2O
© 2004 Key Curriculum Press.
12. Given 8.5 mol of C4H10 and 14.2 mol of O2,
which is the limiting reagent?
2 C4H10
+ 13 O2  8 CO2
8.5 mol C4H10
14.2 mol O2

*
*
8CO2or10H 2O
2molC 4 H10
8CO2or10H 2O
13molO 2
+
= 34 mol CO2 or
42.5 mol H2O
= 8.7 mol CO2 or
10.9 mol H2O
O2 is the limiting reagent!

10 H2O
© 2004 Key Curriculum Press.
13. Octane (C8H18) is a major component of
gasoline. The burning of octane produces carbon
dioxide (CO2), as shown in the equation below.
a)Calculate the molar masses of octane (C8H18) and
carbon dioxide (CO2). Show your calculations and
include units in your answer.
C8H18:
C(8) + H(18) = 12(8) + 1(18) = 114 g/mol
CO2 : C + O(2) = 12 + 16(2) = 44 g/mol
© 2004 Key Curriculum Press.
b) Calculate the amount of CO2, in grams, produced
by the combustion of 100.0 g of octane. Show your
calculations and include units in your answer.
100 g C8H18
= 308.8 g CO2
÷ (114 g/mol)
0.88 mol C8H18
x (44 g/mol)
*
16molCO 2
= 7.02 mol CO2
2molC 8 H18
© 2004 Key Curriculum Press.
The combustion of 100.0 g of propane produces
299.4 g of carbon dioxide. Assume the combustion
of propane produces the same amount of energy
per gram as the combustion of octane.
c) Based on this information and your calculations,
identify which fuel, octane or propane, is less
harmful to the environment. Justify your answer.
Since burning the same amount of propane
produces less CO2 (299.4g instead of 308.8g),
propane is less harmful to the environment than
octane.
© 2004 Key Curriculum Press.
1. The percent yield is
________________________
x 100%
© 2004 Key Curriculum Press.
2 C4H10 + 13 O2
 8 CO2
+ 10 H2O
2.
If you produce 8 moles of H2O from 16
moles of O2 , what is the percent yield?
actual
8molH 2O

theor . 12.3molH 2O = .65 = 65%

16 moles O2
*
10H 2O
13molO 2
= 12.3 mol H2O
© 2004 Key Curriculum Press.
2 C4H10 + 13 O2
 8 CO2
+
10 H2O
3. In an experiment, 1.5 mol C4H10 reacted with
excess oxygen gas (O2). The reaction produced 4.0
mol CO2.What was the percent yield for the reaction?
A. 18.75%
B. 37.5% C. 66.7%
actual 4.0molCO 2

theor . 6.0molCO 2
1.5
moles C4H10 *
8CO2
2C4 H10
D. 85%
= .667 = 66.7%
= 6.0 mol CO2
© 2004 Key Curriculum Press.
4. In your own words, what is the limiting
reactant/reagent?
It is the reactant that ______________________
________________________________________
© 2004 Key Curriculum Press.
5.
Given 2.7 mol of C4H10 and 15 mol of O2,
which is the limiting reagent?
+ 13 O2  8 CO2
2 C4H10
2.7 mol C4H10
15 mol O2

*
*
8CO2or10H 2O
2molC 4 H10
8CO2or10H 2O
13molO 2
+
= 10.8 mol CO2 or
13.5 mol H2O
= 9.2 mol CO2 or
11.5 mol H2O
O2 is the limiting reagent!

10 H2O
© 2004 Key Curriculum Press.
6. Given 3.1 mol of C4H10 and 22 mol of O2 , which
is the limiting reagent?
+ 13 O2  8 CO2
2 C4H10
3.1 mol C4H10
22 mol O2

*
*
8CO2or10H 2O
2molC 4 H10
+
10 H2O
= 12.4 mol CO2 or
15.5 mol H2O
8CO2or10H 2O = 13.5 mol CO or
2
13molO 2
16.9 mol H2O
C4H10 is the limiting reagent!

© 2004 Key Curriculum Press.