Question

I'm working with chemical calculations and need to calculate the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate and I have no idea how to go about doing it.

Answer 1

2 KClO3 ----( Heat with MnO2 Catalyst )----> 2 KCl + 3 O2
Molar mass (molecular weight) of KClO3 is 122.5498 g/mol
Molar mass (molecular weight) of O2 is 31.9989 g/mol
2 moles of KClO3(2*122.5498 g/mol=245.0996)
gives 3 O2(3*31.9989=95.9967 g/mol)
oxygen produced from 75.0 g of potassium chlorate=(95.9967/245.0996)*75=29.3748g
I think this is the way............
sorry if I went wrong.

Answer 2

You could use mass ratios.
\[2KClO_3 \space\xrightarrow{Heat \space with \space MnO_2 \space Catalyst} \space2KCl+3O_2\]
\[mass \space of \space oxygen \space produced \space in \space g = 75g \bigg(\frac{molar \space mass \space of \space\space 3O_2}{molar \space mass \space of \space\space 2KClO_3}\bigg)\]

Answer 3

Okay, I think I get it, but how do you know when to use a catalyst?

Answer 4

Any decent instructor would give you that information if it is important (unless you are taking an extremely specialized course). I assumed hakkunamatata knew what he was talking about. If you search the net, you should be able to find this kind of information:
http://answers.yahoo.com/question/index?qid=20080921023319AAd4GQB