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Answer to Question #56052 in General Chemistry for sue

Question #56052
When heated, KClO3 solid forms solid KCl and O2 gas. A sample of KClO3 is heated and 365 mL of gas with a pressure of 790 mmHg is collected over water, at 26 ∘C. At 26 ∘C, the vapor pressure of water is 25 mmHg: 2(KClO3)(s)→2(KCl)(s)+3(O2)(g)

Part A
What is the pressure of the dry O2 gas?
PO2 = ??

Part B
How many moles of O2 were produced?
n =

Part C
How many grams of KClO3 were reacted?
m = ???
2KClO3(s)→2KCl(s)+3O2(g)
Part A. The total pressure over water consist of the partial pressures of oxygen and water vapour:
P(total) = P(oxygene)+P(vapour);
P(oxygene) = P(total) - P(vapour) = 790 – 25 = 765 mmHg OR 1.007 atm OR 101992 Pa
Part B. The number of moles of oxygen can be calculated using the formula of ideal gas law converting the values to SI units - V = 0.000365 m3 ; T = 299.15 K; R = 8.314 J/(mol*K):

The number of moles using the selected equation is n = 0.015 mole.
Part C. The number of moles of KClO3 is 0.015x2/3 = 0.01 mole
The molar mass of KClO3 is122.55 g/mol
The mass of KClO3 is m = nM = 0.01x122.55 = 1.225 g

Answer: 101992 Pa; 0.015 moles; 1.225 g

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