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Question

In the reaction 2KClO3(s) --> 2KCl(s) + 3O2(g)

How many liters of oxygen gas at STP are produced from 17.4 g of KClO3?

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What is the temperature in oC of 0.742 mole of gas at 2.09 atm and that occupies 9.87 L?

Answer 1

17.4 g of KClO3 (MW=122.5495) is 0.142 mol KClO3.  There are 3 moles O2 produced for every 2 moles KClO3 consumed so the moles of O2 produced would be

 

0.142(3/2) = 0.213 moles O2

 

At STP, 1 mol is 22.4 liters, so 0.213 moles O2 is 0.213(22.4)=4.77 liters.

 

The temperature (°C) of 0.742 mole of gas at 2.09 atm occupying 9.87 L is determined using the ideal gas law as follows:

 

PV = nRT  where R = .08205 atm-liters/(mol-K).  Solving for T gives

 

T=(2.09)(9.87)/[(0.742)(0.08205)]=338.83 K  or 65.68 °C

Answer 2

1 mole of ideal gas at STP occupies 22.4 liters

 

1 mole of KClO₃ = 30.983+35.453+15.994*3 = 114.418 grams

 

17.4 g = 17.4/114.418 = 0.152 moles of KClO₃

0.152 * 22.4 liter = 3.41 liters

 

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PV = nRT

 

T = PV / nR

R = 0.08206 L atm / mol K

 

T = (2.09 atm*9.87 liters) / (0.742 moles * 0.08206)

   = 338.8 K^o

 

0 K^o = -273 C^o

 

338.8 K^o= -273 C^o + 338.8 = 65.8 C^o