A 2.45 gram sample of potassium chlorate is heated an the oxygen gas collected. what is the volume of the oxygen at STP.
Is the oxygen gas itself 2.45 grams or do I have to subtract something from that?
3 answers
No, the KClO3 from which the oxygen is evolved, has a mass of 2.45 g. You need to write the equation, balance it, convert 2.45 g KClO3 to mols, convert mols KClO3 to mols O2 using the coefficients in the balanced equation, then convert mols O2 to liters knowing that 22.4 L is the volume occupied by a mole of a gas at STP. USUALLY, oxygen is collected over water and that must be corrected for the vapor pressure of water; however, your problem doesn't say anything about water; therefore, I asume we don't need to worry about that.Post your work if you get stuck and we can help you through it.
what reactants would there be besides O2?
Heating KClO3 produces oxygen and potassium chloride.
KClO3 + heat ==> KCl + O2
The balanced equation is:
2KClO3 ==> 2KCl + 3O2.
KClO3 + heat ==> KCl + O2
The balanced equation is:
2KClO3 ==> 2KCl + 3O2.
1 answer