Stoichiometry Using Daltons Law of Partial Pressures

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 degrees Celcius and a pressure of 734 torr. Calculate the mass percent of NaClO3 in the original sample. (At 22 degrees Celcius the vapor pressure of water is 19.8 torr.)
 

Answer-                                                 2NaClO3(s)--> 2NaCl(s)+ 3O2(g)

pressure from O2

Ptot = PO2 + PH2O                       734 torr=PO2 +19.8 torr                 PO2 = 714 torr(1 atm/760 torr)= 0.939atm

                                                                                                    V= 57.2mL=0.0572L

                                                                                                     T =22oC + 273= 295K

Solve for Moles

PV=nRT

(0.939atm)(0.0572L)=n(0.0821 atm L mol-1K-1)(295K)

n=0.00222 mol=2.22 x 10-3mol O2

 

2.22 x 10-3mol O2 X

 2 mol NaClO3

X

106.5g NaClO3

 =0.157g
  3 mol O2

1 mol NaClO3
 

% mass=(0.157g/0.8765g) x 100%=18.0%
 
2.30g of an unknown metallic oxide is placed in a 3.00 L flask. The flask is filled with carbon dioxide to a pressure 740. mm Hg at 20. oC. After the reaction, the pressure is then 390. torr at 20.oC. According to the reaction 

XO + CO2 ----> XCO3

 , identify X if it is a group 2A metal (alkaline earth).

 

*****the partial pressure of CO2 that was consumed is

740mmHg-390mmHg= 350mmHg of CO2 used up

convert that to atm
350mmHg (1 atm/760 mmHg) =0.460atm

find the moles

PV=nRT
(0.460atm)( 3.00L)=n 0.0821(293K)
n=0.0574mol CO2 consumed

since it is a 1:1 mol ratio

0.0574 moles of unknown oxide

2.30g/0.0574mol= molar mass =40g/mol

MgO

 

 

 

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