chem help?? GAS PROBLEM??

Question

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8965 g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied 57.2 mL at a temperature of 22°C and a pressure of 738 torr. Calculate the mass percent of NaClO3 in the original sample. (At 22°C the vapor pressure of water is 19.8 torr.)
_____________ %

How would I do this?? I don't know where to start?

Answer 1

Well, you should always start from the reaction equation:
2NaClO3 ==> 2NaCl + 3O2
and also try to do a majority of calculation in mole.
The oxygen gas collected over water occupied 57.2 mL at a temperature of 22°C and a PARTIAL pressure of 718 torr. With ideal gas law, we may figure out the amount of O2:
n = PV/RT = (718/760)*0.0572/(0.08206*295) = 0.00223 (mol)
The reaction equation informs us that 2 mol of NaClO3 may produce 3 mol of O2. Thus 0.00223 mol O2 needs 0.00149 mol of NaClO3 to start with. Since the formula mass of NaClO3 is 106.44 g/mol, 0.00149 mol of NaClO3 is:
(0.00149 mol)*(106.44 g/mol) = 0.158 g.
Hence the requested mass percent of NaClO3 in the original sample is: (0.158 g)/(0.8965 g) = 17.7%.