hee Name Parmer h 1 Generating HydrogGastroo poine Date Section Data Tables (Rec
ID: 694432 • Letter: H
Question
hee Name Parmer h 1 Generating HydrogGastroo poine Date Section Data Tables (Record data from all of the groups in your Lab Section) 110 points) Group Number Atmospheric Pressure in the La mmHg 26.3C Temperature of WerMass of MagnesiumVlume of Hydrogen Group Number (Kelvin) Metal (grams) Gas (ml.) 248.35 10 12 13 14 15 16 Average 298.35 k 1. Please use the average values in the Data Table to calculate the molar volume for the hydro- gen gas at STP in units of L/mol. Also calculate the % ERROR (theoretical molar volume of a gas at STP is 22.4 L/mol). Sbow all calculations on an attached sheet; also include correct units and significant figures. [25 points 143Explanation / Answer
According to Ideal gas equation,
PV = nRT
P = pressure of the gas = 759 mmHg = 759/760 = 0.9986 atm
V = volume of the gas = 41.3 mL = 41.3 * 10^3 L
n = m/M
m = mass of the gas = 0.041 g
M = molar mass
T = absolute temperatue = 298.35 K
PV = m/M RT
0.9986 * 41.3 * 10^3 = 0.041/M*0.0821 * 298.35
M = 2.43 g/mol
Molar volume = molar mass/molar density
V = 2.43/0.08988 = 27.036 L/mol
Percentage of error = eXperimental value/Theoritical value *100
% error = 27.036/22.4*100 = 120.944%
2 HgO + Heat ----> 2 Hg + O2
Volume of O2 gas = 125 mL = 0.125 L
pressure of O2 gas = P gas - p H2O
PO2 = 763 mm Hg - 22.4 mm Hg = 740.6 mmHg = 740.6/760 = 0.9745 atm
Temperature = 24 oC = 273.15 + 24 = 296.15K
From ideal gas equation,
PV = nRT
0.9745 * 0.125 = n * 0.0821 * 296.15
n = 5 * 10^-3 moles
2 moles of HgO produces 1 mole of O2
2 * 216.6 g of HgO produces 32 g of O2
2.154 g of HgO produces 32/216.6 * 2.154 g of O2
= 0.3182 g
n = m/M
M = molar mass of O2
m = mass of O2
5 * 10^-3 = 0.3182/M
M = molar mass of O2 = 63.64
Molar volume = molar mass/molar density
V = 63.64/1.4286 = 44.547 L/mol
% error = 44.547/22.4*100 = 198.8%
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