Asked by Anonymous
The weight of kclo3 which thermal decomposition o2 gas sufficient for the combustion of 1.12 litres of ethlene gas at STPis
Answers
Answered by
DrBob222
C2H4 + 3O2 ==> 2CO2 + 2H2O
mols 1.12 L ethylene = 1.12L/22.4L = approx 0.05 but you should confirm that.
So what mass KClO3 will produce 0.05 mols O2 gas?
2KClO3 ==> 2KCl + 3O2
0.05 mols O2 x (2 mols KClO3/3 mols O2) = 0.05 x 2/3 = approx 0.03 but you should get an accurate answer.
Then mass KClO3 = mols KClO3 x molar mass KClO3.
mols 1.12 L ethylene = 1.12L/22.4L = approx 0.05 but you should confirm that.
So what mass KClO3 will produce 0.05 mols O2 gas?
2KClO3 ==> 2KCl + 3O2
0.05 mols O2 x (2 mols KClO3/3 mols O2) = 0.05 x 2/3 = approx 0.03 but you should get an accurate answer.
Then mass KClO3 = mols KClO3 x molar mass KClO3.
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