Experiment 8 Report Sheet KI (cat) 2H20(aq) 020g) 2 H202(aq) Data NCLUDE UNITS T
ID: 505273 • Letter: E
Question
Experiment 8 Report Sheet KI (cat) 2H20(aq) 020g) 2 H202(aq) Data NCLUDE UNITS T Barometric Pressure: Ida13 Mass of 10 mL grad cylinder with 5 mL A H202 solution B Mass of empty 10 mL grad cylinder C Initial gas volume reading D Final gas volume reading Vapor pressure of water F Calculations: INCLUDE UNITS AND SHow WORK (attach scratch paper or use back side of the page)!!! G Mass of H202 solution used H Volume of gas collected, in mL I Volume of gas collected, in L J Partial pressure of O2, in torr K Partial pressure of 02, in atm L Temperature (K) M Number of moles of 02 formed N Number of moles of Hizon consumed Mass of H202 consumed o (molar mass 34,02 g/mol) P Mass percent Hzoa in solution Q Average mass 96 Hzoz 8-7Explanation / Answer
The given reaction is
2 H2O2 (aq) ------> 2 H2O (aq) + O2 (g)
As per the balanced stoichiometric reaction,
2 moles H2O2 = 1 mole O2 …..(1)
Barometric pressure = 765.54 torr
Trial 1
Trail 2
A
Mass of 10 mL graduated cylinder with 5 mL H2O2 (g)
64.16
65.03
B
Mass of empty 10 mL graduated cylinder (g)
60.01
60.04
C
Initial gas volume reading (mL)
32
34
D
Final gas volume reading (mL)
0
0
E
Temperature (C)
30
30
F
Vapor pressure of water (torr) (obtained from internet sources)
31.8
Calculations:
Trial 1
Trial 2
G
Mass of H2O2 solution used (g) (A – B)
4.15
4.99
H
Volume of gas collected, in mL (C – D)
32
34
I
Volume of gas collected, in L (H/1000)
0.032
0.034
J
Partial pressure of O2 in torr (Barometric pressure – F)
733.74
K
Partial pressure of O2 in atm (J *0.001316)
0.9656
L
Temperature (K) (E + 273)
303
M
Number of moles of O2 formed
0.001244 (check calculation below)
0.001321
N
Number of moles of H2O2 consumed
0.002488
0.002642
O
Mass of H2O2 consumed (molar mass = 34.02 g/mol)
0.08464
0.08988
P
Mass percent of H2O2 in solution (O/G)*100
2.0395
1.8012
Q
Average mass percent H2O2
(2.0395 + 1.8012)/2 = 3.8407/2 = 1.92035 1.920
Moles of O2 formed:
Use the ideal gas law, P*V = n*R*T where n = moles of O2 formed.
(0.9656 atm)*(0.032 L) = n*(0.082 L-atm/mol.K)*(303 K)
===> n = 0.001244
Moles of H2O2 consumed:
As per the balanced equation above,
1 mole O2 = 2 moles H2O2
===> 0.001244 mole O2 = (0.001244 mole O2)*(2 mole H2O2/1 mole O2) = 0.002488
Trial 1
Trail 2
A
Mass of 10 mL graduated cylinder with 5 mL H2O2 (g)
64.16
65.03
B
Mass of empty 10 mL graduated cylinder (g)
60.01
60.04
C
Initial gas volume reading (mL)
32
34
D
Final gas volume reading (mL)
0
0
E
Temperature (C)
30
30
F
Vapor pressure of water (torr) (obtained from internet sources)
31.8
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