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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.)

Respuesta :

Willchemistry
Molar mass :

KClO₃ = 122.55 g/mol
O₂ = 32 g/mol

2 KClO₃ = 2 KCl + 3 O₂

2 x 122.55 g KClO₃ ----------> 3 x 32 g O₂
10.0 g KClO₃ ------------------> ?

Mass of O₂ = ( 10.0 x 3 x 32 ) / ( 2 x 122.55 )

Mass of O₂ = 960 / 245.1

Mass of O₂ = 3.9167 g

number of moles O₂ => 3.9167 / 32 = 0.1223 moles 

1 mol -------------- 22.4 L ( at STP)
0.1223 moles ---- ?

V = 0.1223 x 22.4 / 1

V = 2.739 L

hope this helps!

Your answer is 2.74.

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