You are given 1.446 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl and O2,
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),
and 355 mL of O2 is collected over water at 26 °C. The total pressure of the gases in the collection flask is 745 torr. What is the weight percentage of KClO3 in the sample?
The formula weight of KClO3 is 122.55 g/mol. The vapor pressure of water at 26 °C is 25.2 torr.
what is the weight percentage of KCIO3 in this sample? I have no idea how to do this..this question is due in 3 hours please help me?
2017-04-08 12:17 am
回答 (2)
2017-04-08 12:32 am
Consider the oxygen gas collected :
Pressure , P = (745 - 25.2) torr = 719.8 torr = 719.8/760 atm
Volume, V = 355 mL = 0.355 L
No. of moles, n = ? mol
Gas constant, R = 0.08206 L atm / (mol K)
Absolute temperature, T = (273 + 26) K = 299 K
Gas law : PV = nRT
Then, n = PV/(RT)
No. of moles of O₂ formed, n = (719.8/760) × 0.355 / (0.08206 × 299) mol = 0.0137 mol
2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
Mole ratio KClO₃ : O₂ = 2 : 3
No. of moles of KClO₃ in the mixture = (0.0137 mol) × (2/3) = 0.009133 mol
Formula weight of KClO₃ = 122.55 g/mol
Weight of KClO₃ in the mixture = (0.009133 mol) × (122.55 g/mol) = 1.119 g
Weight percentage of KClO₃ = (1.119/1.446) × 100(%) = 77.39%
Pressure , P = (745 - 25.2) torr = 719.8 torr = 719.8/760 atm
Volume, V = 355 mL = 0.355 L
No. of moles, n = ? mol
Gas constant, R = 0.08206 L atm / (mol K)
Absolute temperature, T = (273 + 26) K = 299 K
Gas law : PV = nRT
Then, n = PV/(RT)
No. of moles of O₂ formed, n = (719.8/760) × 0.355 / (0.08206 × 299) mol = 0.0137 mol
2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
Mole ratio KClO₃ : O₂ = 2 : 3
No. of moles of KClO₃ in the mixture = (0.0137 mol) × (2/3) = 0.009133 mol
Formula weight of KClO₃ = 122.55 g/mol
Weight of KClO₃ in the mixture = (0.009133 mol) × (122.55 g/mol) = 1.119 g
Weight percentage of KClO₃ = (1.119/1.446) × 100(%) = 77.39%
2017-04-08 12:36 am
...
find moles of O2
PV = nRT
(745/760) atm (0.355 L) = n (0.08206 L*atm/mol*K) (299 K)
n = 0.01418 mol
find mass of KClO3
. . . 1.159g
. . 2KClO3. . → , , 2KCl . . + . . 3O2
(122.55g/mol)
0.009455mol . . . . . . . . . . . . . 0.01418mol
. . . ↑|________2 to 3__________|↓
find percentage of KClO3 in the mix
X/100% = KClO3 / mix = 1.159 g / 1.446 g = 0.8013
X = 80.13%
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
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find moles of O2
PV = nRT
(745/760) atm (0.355 L) = n (0.08206 L*atm/mol*K) (299 K)
n = 0.01418 mol
find mass of KClO3
. . . 1.159g
. . 2KClO3. . → , , 2KCl . . + . . 3O2
(122.55g/mol)
0.009455mol . . . . . . . . . . . . . 0.01418mol
. . . ↑|________2 to 3__________|↓
find percentage of KClO3 in the mix
X/100% = KClO3 / mix = 1.159 g / 1.446 g = 0.8013
X = 80.13%
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
You may have to wait an hour to award BA.
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