Decomposition of potassium chlorate (kclo3) produces potassium chloride (kcl) and pure oxygen (o2). the balanced equation for the reaction is as follows.
mc017-1.jpg
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.)

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!

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !