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Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2). The balanced equation for the reaction is as follows.
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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)
0.914 L
1.83 L
2.74 L
3.66 L

Respuesta :

taskmasters

1.    The balanced reaction would be:

2KClO3 --> 2KCl + 3O2


We are given the amount of potassium chlorate for the reaction. This will be the starting point of our calculation. 

10.0 g KClO3 ( 1 mol KClO3 / 122.55 g KClO3) (3 mol O2 / 2 mol KClO3) = 0.1224 mol O2

V = nRT/P =0.1224 mol O2 x 273 K x 0.08206 atm L/mol K / 1 atm
V=2.74 L

Answer:

third option 2.74L

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