I need help with the bottom portion of the paper, which is the all the calculati

Question

I need help with the bottom portion of the paper, which is the all the calculations (Please show all work for each). The top half is data from the experiment perfomred.

The objective of the experiment was to, 1. To use gas data to determine the value of the molar volume of oxygen at standard temperature and pressure (273.15 K and 1 atm) 2. To use stoichiometry of a decompostion reaction to determine the percent KClO3 in a sample.

2KClO3 ----> 2KCl +3O2 (MnO2 + Heat was added to make the solution)

Trial 2 k Mass of empty tube after initial heating .222. 21-218 Mass of tube plus sample before heating Mass of tube plus sample after heating 013 iss m volume of water displaced 22 22 Water temperature (C) Barometric pressure CALCULATIONS Show your tions 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 Calculated volume of O, sample at STP Calculated molar volume of O, at STP Average molar volume Percent error in molar volume

Explanation / Answer

2KClO3 ----> 2KCl +3O2

mass of Oxygen:

Volume of water displaced = volume fo Oxygen

so, For trial 1:

Volume of O2 = 155 mL, at P = 743.24 mm Hg, for water at 22°C = 19.827 mm Hg

total Pressure of Gas = 743.24-19.827 = 723.413 mm Hg of O2

T = 22°C = 22+273 = 295 K

PV = nRT

n = PV/(RT)

n = (723.413 )(155*10^-3)/(62.3*295) = 0.006101 mol of O2

mass = mol*MW = 0.006101*32 = 0.195232 g of O2

moles of O2 generated --> 0.006101 mol of O2

vapore pressure of water --> 19.827 mm Hg at 22°C

Pressure of O2 -->  743.24-19.827 = 723.413 mm Hg

T of O2 --> 22°C = 295 K

volume of O2 at stp...

1 mol of gas = 22.4 L at STP

0.006101 mol = 0.006101 *22.4 = 0.1366624 L of O2 at STP

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