Trial 1 Trial 2 Mass of empty tube after initial heating 0.0219 Mass of tube plu

Question

Trial 1 Trial 2 Mass of empty tube after initial heating 0.0219 Mass of tube plus sample before heating 3 lao2 Mass of tube plus sample after heating Volume of water displaced Water temperature (C) Barometric pressure ll os09 1t lo029 1.5 09 240, on 230.090 21.0( 23.0°C 242 2o tem CALCULATIONS Show your caleulations 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 ofo Temperature of O, (K) 3 0.30 Calculated volume of O, sample at STP Calculated molar volume of O, at STP Average molar volume Percent error in molar volume

Explanation / Answer

Decomposition of KClO3

2KClO3(s) --> 2KCl(s) + 3O2(g)

From the abvove data,

Trial 1,

mass of KClO3 sample = 1.581 g

mass of O2 generated = 0.268 g

moles O2 generated = 0.268 g/32 g/mol = 0.0084 mol

Volume of water displaced = 240 ml = 0.240 L

Temperature of water = 24 oC + 273 = 297 K

Vapor pressure of water at 24 oC = 22.4 torr

Pressure of O2 = 743.26 - 22.4 = 719.86 torr

                         = 719.86/760 = 0.95 atm

Volume of O2 at STP = 0.95 atm x 0.240 L x 273.15 K/1 atm x 297 K = 0.21 L

molar volume at STP = 0.21 L/0.0084 mol = 25 L/mol

Trial 2,

mass of KClO3 sample = 1.520 g

mass of O2 generated = 0.303 g

moles O2 generated = 0.303 g/32 g/mol = 0.0095 mol

Volume of water displaced = 230 ml = 0.230 L

Temperature of water = 23 oC + 273 = 296 K

Vapor pressure of water at 24 oC = 21.1 torr

Pressure of O2 = 742.26 - 21.1 = 721.16 torr

                         = 721.16/760 = 0.95 atm

Volume of O2 at STP = 0.95 atm x 0.230 L x 273.15 K/1 atm x 296 K = 0.202 L

molar volume at STP = 0.202 L/0.0095 mol = 21.3 L/mol

Average molar volume of O2 = 23.2 L/mol

Percent error in molar volume = (23.3 - 22.4) x 100/22.4 = 4.02%

Percent KClO3 in sample

Trial 1,

moles O2 generated = 0.268 g/32 g/mol = 0.0084 mol

mass KClO3 = 0.0084 mol x 122.55 g/mol = 1.03 g

mass% KClO3 in sample = 1.03g x 100/1.581g = 65.15%

Trial 2,

moles O2 generated = 0.268 g/32 g/mol = 0.0084 mol

mass KClO3 = 0.0095 mol x 122.55 g/mol = 1.164 g

mass% KClO3 in sample = 1.164g x 100/1.520g = 76.58%

average mass% KClO3 in sample = 70.86%

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