At STP potassium chlorate (KClO3) decomposes to produce solid potassium chloride (KCL) and oxygen gas (O2) according to the balanced chemical equation:
2KClO3(s)->2KCl(s)+3O2(g)
What volume of oxygen gas, measured at 40 degrees celsius and 85.4 kPa, will be produced when 13.5 g of potassium chlorate is decomposed? Answer is 5 L. what gas law should i use?
3 answers
PV = nRT
From equation 1 mole KClO3 => 3/2 mole O2 at STP x 22.4L/mole = 33.6L at STP.
Given 13.5g KClO3/(122g/mol)=> 3.73L O2 at STP. Adjusting for 40-deg C and 85.4Kpa (~641mm) ...
Vol@40C&641mm
= 3.73(760/641)(313/273)L
= 5.07 Liters
Given 13.5g KClO3/(122g/mol)=> 3.73L O2 at STP. Adjusting for 40-deg C and 85.4Kpa (~641mm) ...
Vol@40C&641mm
= 3.73(760/641)(313/273)L
= 5.07 Liters
If you can calculate moles O2 ~0.167mole O2 from 13.5g KClO2 and use 313K & (641/760)atm then, PV=nRT is a shorter route.
1 answer