Transcript Document

1. Your friend Becky is sanitizing her pool by adding Ca(ClO)2 to the water. She wants to
go swimming, but you are afraid she may have added to much Ca(ClO)2 and will die if
she swims in the pool. You know that it is dangerous to swim in a pool with more
than a .05 Molar concentration of ClO- in the water. If Becky’s swimming pool holds
275 gallons of water and she added 75 grams of Ca(ClO)2 to the water, calculate the
molarity and find out if it is safe to swim in.
(Hint: 1 gallon = 3.79 liters)
2. 150 grams of Hydrogen gas is reacted with 1500g O2 gas, to form H2O. How many
grams of water will be made?
2H(g) + O2(g) =2H2O(l)
3. Balance the following equation and solve for how many grams of C02 are formed if 10
grams of CO are reacted with excess oxygen gas.
CO+ 02 = CO2
4. Given the following balanced equation, how much C6H5Br will be produced when 30g
C6H6 reacts with 65g of Br2?
C6H5 + Br2 = C6H5Br + HBr
5. Balance the following equation and then use it to find out how many grams of NaI are
needed to remove 1.3 Kg of O3?
6. O3 + NAI + H2O = O2 + I2 NaOH
(Hint: 1Kg = 1000g)
6.)
2C6H12(l) + 5O2(g) = 2H2C6H8O4(l) + 2H2O(g)
Use the provided balanced equation to answer the question. If the limiting reactant in
this equation is C6H12 and 25g of it are used, what is the theatrical yield of H2C6H8O4?
b) If after performing the experiment 33.5 g of H2C6H8O4 was made, what is the
percent yield?
7.) 30g of an unknown organic compound are combusted, leaving you with 733g of
CO2 and 30g of water. What is the empirical formula of the compound you just
combusted?
8.)
2NaOH + CO2 +Na2CO3 + H2O
When 1.85 mol of NaOH reacts with 1 mol Co2 which is the limiting reactant?
b.) How many moles of the excess reactant are left?
Answer key to the Stoich Study
Guide.
1
275 Gallons
H20
= 1045.25 L H2O
3.79 L
1 gallon
75 g Ca(ClO)2
1 mole Ca(ClO)2
141.1g
Moles solute
M =
Liters solution
2 moles ClO
51.5g
1mole
Ca(ClO)2
1 mole ClO
=
51.5g
1042.25 L
=
= 54.748g ClO
.05M
2
150g H2
1 mol H2
=75 mol H2
2.00g H2
1500g O2
1 mol O2
2 mol O
32.0g O2
1 mol 02
=94 moles O
H2 is the limiting reactant
75 mol H2
2 mol H2O
2 mol H2
18.0g H2O
1 mol H2O
=1400g H2O
2Co + O2 = 2CO2
3.
10g CO
4
30g C6H6
1 mole Co
2 mole CO2
28 g
2 mol CO
1 mole C6H6
65g Br2
1 mole C6H6
78g
1 mol C6H5Br
1 mole Br2
1 mole Br2
160g
1 mol C6H5Br
44g
= 15.7g CO2
1 mole CO2
157g
= 60.4g C6H5Br
1 mole C6H5Br
157g
= 63.8g C6H5Br
1 mole C6H5Br
C6H6 is the limiting reactant, 60.4g of C6H5Br is made
O3 + 2NaI + H20 = O2 + I2 + 2NaOH
5
1.3 Kg O3
1000g
1Kg
1 mole O3
48g
2 mole NaI
150g
1 mole O3
1 mol NaI
= 812.5g C6H5Br
6
25g C6H12
1 mol C6H12
84g
1 mol H2C6H8O4
146g H2C6H8O4
2 mol C6H12
1 mol H2C6H8O4
= 43.5g H2C6H8O4 = theoretical Yield
b.
actual
Theoretical
X 100%
33.5g
43.5
X 100%
=77%
percent
yield.
7
733g CO2
1 mol CO2
=16.6 mol CO2
44g
30g H2O
1 mol H2O
=16.6 mol
H2O
18g
2 mol H
16.6 mol H2O
=33.3 mol H
1 mol H2O
16.6 mol C
12g
= 66.6% C
1 mole C
33.3 mol H
300 g
1g
= 11.3% C
1 mole H
300 g
100- (66+11.3)= 22.1% O
.221 O
66.3g O
300g
=66.3g O
1 mole O
=4.125 mol O
16.6g
÷
C
H
16.6
33.3
4.13
4
÷
4.13
8
O
÷
4.13
4.13
1
Formula is C4H8O
8
1.85 mol NaOH
1 mole
H2O
1 mol
NaOH
1 mol CO2
1 mole
H2O
1 mol
CO2
18g
H2O
1mol H2O
18g
H2O
1mol H2O
=16.65g
H2O
=18g
H2O
NaOH is the limiting reactant.
b.
16.65g H2O
1 mol
H2O
1 mole
CO2
18g
H2O
1mol H2O
1 mole CO2 - .925 moles CO2 =
=.925 moles
CO2 used
.075 mol excess
reactant