Question

Chemistry 201 Experiment 10: Determination of the ideal Gas Constant Data and Observations Data T rial Trial #2 Trial 43 Mass of sulfamic acid 0.5782 Mass of sodium nitrite 0.6114 41.59 40.39 40.41 190 19.0 190 2.198 2.198 2.198 Volume of water displaced by N (9) Water temperature Vapor pressure of water from CAC Handbook) Atmospheric pressure (from barometer) Pressure of nitrogen gas Moles NaNO, used per trial (calculation below) Moles HSO NH, used per trial (calculation below) kPa 92.41 92.41 92.41 kPa mol mol Number of moles of NaNO, used in each trial (10.0 mL of 40.0 ml, unless modified) - Number of moles of HSONH, used in each trial (25.0 mL of 100.0 ml, unless modified) =

Fill in the blank

Answer 1

The givenn chemical reaction is

NaNO2 + HSO3NH2 ------------> NaHSO4 + N2 + H2O

That means 1 mole of N2 is produced from the reaction of 1 mole of NaNO2 and 1 mole of HSO3NH2

Points to remember-

• A gas when added to water, displace water of equal to the volume of its own volume.
• Value of gas constant R = 8.314 L kPa K−1 mol−1

Now given

Trial 1-

Volume eof water dispaced by N2 = Volume of N2 gas itself = 41.59 mL = 0.04159 L

Temp of water = Temp of N2 gas = 19oC = 19 + 273 = 292‬ K

Pressure of N2 = Total pressure - Pressure of water vapor = 92.41 - 2.198 = 90.212 kPa

That means moles of N2 produced (n) can nbe calculataed from Ideal gas equation-

PN2 * VN2 = nRT

90.212 kPa * 0.04159 L = n * (8.314 L kPa K−1 mol−1)* 292‬ K

n = [90.212 kPa * 0.04159 L] / [(8.314 L kPa K−1 mol−1)* 292‬ K]

= [90.212 * 0.04159 ] / [(8.314 )* 292‬ ] (kPa * L * K. mol / K. L. kPa)

= 0.0015 moles

That means moles of N2 produced = 0.0015 moles

So moles of NaNO2 used in reaction = 0.0015 moles

And moles of HSO3NH2 used in reaction = 0.0015 moles

Similarly-

Trial 2-

Volume eof water dispaced by N2 = Volume of N2 gas itself = 40.39 mL = 0.04039 L

Temp of water = Temp of N2 gas = 19oC = 19 + 273 = 292‬ K

Pressure of N2 = Total pressure - Pressure of water vapor = 92.41 - 2.198 = 90.212 kPa

That means moles of N2 produced (n) can nbe calculataed from Ideal gas equation-

PN2 * VN2 = nRT

90.212 kPa * 0.04039 L = n * (8.314 L kPa K−1 mol−1)* 292‬ K

n = [90.212 kPa * 0.04039 L] / [(8.314 L kPa K−1 mol−1)* 292‬ K]

= [90.212 * 0.04039 ] / [(8.314 )* 292‬ ] (kPa * L * K. mol / K. L. kPa)

= 0.0015 moles

That means moles of N2 produced = 0.0015 moles

So moles of NaNO2 used in reaction = 0.0015 moles

And moles of HSO3NH2 used in reaction = 0.0015 moles

Trial 3-

Volume eof water dispaced by N2 = Volume of N2 gas itself = 40.41 mL = 0.04041 L

Temp of water = Temp of N2 gas = 19oC = 19 + 273 = 292‬ K

Pressure of N2 = Total pressure - Pressure of water vapor = 92.41 - 2.198 = 90.212 kPa

That means moles of N2 produced (n) can nbe calculataed from Ideal gas equation-

PN2 * VN2 = nRT

90.212 kPa * 0.04041 L = n * (8.314 L kPa K−1 mol−1)* 292‬ K

n = [90.212 kPa * 0.04041 L] / [(8.314 L kPa K−1 mol−1)* 292‬ K]

= [90.212 * 0.04041 ] / [(8.314 )* 292‬ ] (kPa * L * K. mol / K. L. kPa)

= 0.0015 moles

That means moles of N2 produced = 0.0015 moles

So moles of NaNO2 used in reaction = 0.0015 moles

And moles of HSO3NH2 used in reaction = 0.0015 moles