Identify some of the sources of error that may explain why your calculated value
ID: 1049882 • Letter: I
Question
Identify some of the sources of error that may explain why your calculated value of the molar volume is different than that of an ideal gas. Sodium nitrate decomposes into sodium nitrite and oxygen gas with heat. When 0.123 g of sodium nitrate was heated, it produced 162 mL of oxygen collected over water. The room temperature and pressure are 22.5 degree C and 758 torr, respectively. The water over which the gas was collected had a temperature of 20.0 degree C. Calculate the theoretical amount of oxygen in grants which should have been produced. What is the percent yield? Theoretical yield oxygen: Actual yield oxygen: Percent yield: %Explanation / Answer
Q1.
error in molar volume for ideal gas...
1 - incorrect handliung of material
2- incorrect operation/measurement of volume/temperature
3- absolute pressure is not used
4- absolute temperautre was not converted
5- this is real gas and can't be modeled with ideal gas
Q2.
NaNO3 --> NaNO2 + 1/2O2
m = 0.123 g of NaNO3
mol of NaNO3 = mass/MW =0.123 / 84.9947 = 0.001447 mol of NanO3
then, since ratio is 1:1/2 then
0.001447 mol --> 1/2*0.001447 = 0.000723 mol of O2 expected
V = 16.2 mL of O2 over water
Patm = 758 torr Tabm = 22.5°C = 295.5 K
P°(20°C) = 17.5424
Ptotal = PAtm + Pvapor
Pgas = PTotal - Pvapor= 758 - 17.5424 = 740.46 torr = 740.4/760 = 0.974210 atm
R = 0.082 atmL/mol K
calcualte mol
PV = nRT
n = PV/(RT)
n = (0.974210)(16.2*10^-3)/(0.082*295.5 ) = 6.51322*10^-4 mol
theoretical mol of O2 = 0.000723 mol
%yield = actual/real * 100% = ( 6.51322*10^-4)/0.000723 *100 = 90.0860
mass of O2 = mol*MW = 6.51322*10^-4)*32 = 0.02084230 g