c) A sample ofgas has a volume of25ILat 263 (Cand 7562 minllg. what would its vo
ID: 1039984 • Letter: C
Question
c) A sample ofgas has a volume of25ILat 263 (Cand 7562 minllg. what would its volume be at STP Oxygen gas is sometimes made in the laboratory by heating solid potassium chlorate, KCIO(s). The other product is potassium chloride. What volume of oxygen gas at 21.6 "C and 763 mmHg will be obtained from 29.6 g of potassium chlorate? a) What is the partial pressure of each gas in a mixture of 2.43 mol nitrogen gas and 3,07 mol oxygen gas in a 5.00-L container at 298 K? b) What is the total pressure of the mixture of gases in part (a)? (Assume ideal behavior)Explanation / Answer
5.
Given conditions STP conditions
V1 = 2.51L V2 =
P1 = 756.2mmHg P2 = 760mmHg
T1 = 26.3 + 273 = 299.3K T2 = 273K
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1P2
= 756.2*2.51*273/299.3*760 = 2.27L
6. 2KClO3-------------> 2KCl + 3O2
2 moles of KClO3 decomposes to gives 3 moles of O2
2*122.5g of KClO3 decomposes to gives 3 moles of O2
29.6g of KClO3 decomposes to gives = 3*29.6/2*122.5 = 0.36 moles of O2
n = 0.36moles
T = 21.6+273 = 294.6K
P = 763/760 = 1atm
PV = nRT
V = nRT/P
= 0.36*0.0821*294.6/1 = 8.7L
7. no of moles of N2 (nN2) = 2.43moles
no of moles of O2(nO2) = 3.07moles
total no of moles = nN2 + nO2 = 2.43+3.07 = 5.5moles
V = 5L
T = 298K
PV = nRT
P = nRT/V
= 5.5*0.0821*298/5 = 26.9atm
The total pressure(PT) = 26.9atm
mole fraction of nitrogen (Xn2) = nN2/nN2 + nO2
= 2.43/5.5 = 0.442
mole fraction of oxygen (XO2) = nO2/nN2 + nO2
= 3.07/5.5 = 0.558
partial pressure of nitrogen (PN2) = XN2* PT
= 0.442*26.9 = 11.89atm
partial pressure of oxygen (PO2) = XO2*PT
= 0.558*26.9 = 15 atm
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