DATA Apparatus UNKNOWN NUMBER: ? Approval Trial 1 Trial 2 Mass of empty tube aft

Question

DATA Apparatus UNKNOWN NUMBER: ? Approval Trial 1 Trial 2 Mass of empty tube after initial hearing Mas of ube plus sample before heatng/e 9053 1.38 noskg-- 2.55 Mass oftube plus sample after heating Volume of water displaced Water temperature (C) Barometric pressure go oc 43.24 tor CALCULATIONS Show your calculations in the space provided on the next page. Molar Volume of Oxygen: Mass of oxygen generated Moles of O, generated Vapor pressure of water Pressure of O, Temperature of O, (K) Calculated volume of 02 sample at STP- Calculated molar volume of O, at STP_ Average molar volume Percent error in molar volume

Explanation / Answer

Molar volume calculation

Trial 1,

mass of O2 generated = 16.908 - 16.666 = 0.242 g

moles of O2 = 0.242 g/32 g/mol = 0.0076 mol

Volume of gas (V1) = 190 ml = 0.190 L

Temperature (T1) = 24 oC + 273 = 297 K

vapor pressure of water at 24 oC = 22.4 torr

partial pressure of O2 (P1) = 743.24 - 22.4 = 720.84 torr

                                            = 720.84/760 = 0.95 atm

Volume of O2 at STP = P1V1T2/T1P2

with P2 and T2 be pressure and temperature at STP

                                   = 0.95 x 0.190 x 273.15/297 x 1 = 0.166 L

molar volume of O2 = 0.166 L/0.0076 mol = 21.843 L/mol

Trial 2,

mass of O2 generated = 0.330 g

moles of O2 = 0.330 g/32 g/mol = 0.0103 mol

Volume of gas (V1) = 233 ml = 0.233 L

Temperature (T1) = 23.5 oC + 273 = 296.5 K

vapor pressure of water at 23.5 oC = 21.75 torr

partial pressure of O2 (P1) = 743.24 - 21.75 = 721.49 torr

                                            = 721.49/760 = 0.95 atm

Volume of O2 at STP = P1V1T2/T1P2

with P2 and T2 be pressure and temperature at STP

                                   = 0.95 x 0.233 x 273.15/296.5 x 1 = 0.204 L

molar volume of O2 = 0.204 L/0.0103 mol = 19.806 L/mol

Average molar volume of O2 = 20.8245 L/mol

percent error in molar volume = (22.4 - 20.8245) x 100/22.4 = 7.03%

Percent KClO3 in the sample

2KClO3 --> 2KCl + 3O2

Trial 1,

moles O2 = 0.0076 mol

mass KClO3 = 0.0076 mol x 2 x 122.55 g/mol/3 = 0.9314 g

percent KClO3 in sample = 0.9314 g x 100/1.485 g = 62.72%

Trial 2,

moles O2 = 0.0103 mol

mass KClO3 = 0.0103 mol x 2 x 122.55 g/mol/3 = 0.84151 g

percent KClO3 in sample = 0.84151 g x 100/1.989 g = 42.31%

average %KClO3 in sample = 52.515%

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