I worked on these two chem questions all night and I still couldn't solve them. Help PLEASE!!!!?

Question

1. A sample of oxygen gas is collected over water at 25 C and a total pressure of 641 torr. The volume of gas collected is 500 mL. What mass of oxygen is collected? (At 25 C the vapor pressure of water is 23.8 torr.)

2. At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8765 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 734 torr. Calculate the mass percent of NaCLO3 in the original sample. (At 22 C the vapor pressure of water is 19.8 torr.)

*The three after the O is a subscript.

Show all work please. Thank you SOOO much to everyone who helps!

Answer 1

1. A sample of oxygen gas is collected over water at 25 C and a total pressure of 641 torr. The volume of gas collected is 500 mL.  (At 25 C the pressure of the oxygen is 641 Torr - 23.8 torr = 617.2 Torr)

first find moles of oxygen:

PV=nRT

(617.2T0rr)(0.500Litres) = n (62.34 Torr-Litres/mol-K)(298K)

n= 0.0166 moles of oxygen

What mass of oxygen is collected?

0.0166 moles @ 32g/mole =

your answer:  0.531 grams of oxygen

======================================...

2. A 0.8765 g sample  produced  57.2 mL oxygen gas collected over water at 22 C and  734 torr.  (At 22 C the  pressure of oxygen is 734Torr -19.8 torr = 714.2 Torr)

first find moles of oxygen:

PV=nRT

(714.2Torr)(0.0572Litres) = n (62.34 Torr-Litres/mol-K)(295K)

n= 0.00222 moles of oxygen

now use the ratio: 2 NaClO3 --> 3 O2 to find moles of NaClO3:

0.00222 moles @ 2 moles NaClO3 / 3 moles O2 = 0.00148 moles NaClO3

now find the grams of NaClO3 using its molar mass:

0.00148 moles NaClO3 @ 106.44 g/mol = 0.1576 grams of NaClO3

last, find  the mass percent of NaCLO3 in the original sample:

0.1576 grams / 0.8765 g sample timnes 100 = 17.98%

your answer(3 sigfigs) : 18.0%

Answer 2

pressure O2 = 641 - 23.8 = 617.2 torr

p = 617.2 / 760 = 0.812 atm

T = 25 + 273 = 298 K

V = 0.500 L

n = pV / RT = 0.812 x 0.500 / 0.0821 x 298 =  0.0166 moles

mass = 0.0166 mol x 32 g/mol = 0.531 g

2 NaClO3 >> 2 NaCl + 3 O2

p = 734 - 19.8 = 714.2 torr =>  0.940 atm

V = 0.0572 L

T = 22 + 273 = 295 K

n = pV / RT = 0.940  x 0.0572 / 0.0821 x 295 = 0.00222

Moles NaClO3 = 0.00222 x 2 / 3 = 0.00148

Mass = 0.00148 x 106.44 g/mol = 0.157 g

% = 0.157 x 100 / 0.8765 = 17.97