CHEM 1211 Lab Mamual Revised 05/2017 Ideal Gas Law, PV = nRT Name Section Partne
ID: 702134 • Letter: C
Question
CHEM 1211 Lab Mamual Revised 05/2017 Ideal Gas Law, PV = nRT Name Section Partner Instructor (Unless otherwise noted, 2 pts. for each blank) Molarity of HCI (M) (Ipt. each): Volume of HC1 (mL): Mass of Mg (gram): *Volume of gas before placing ino-Im Data, Trial 31-4 mL 0.0369 Data, Trial 2 31.1mL Equalization Chamber (mL) (1 pt) (This value NOT used in cakulatioes b Volume of gas after placing 31.5ml 31.5m1 in Equalization Chamber (mL): Barometric Pressure (mm Hg): Temperature (): Write the balanced overall equation for the reaction of magnesium and HCI to form hydrogen 205 20.5 gas and the magnesium salt (2pts): Mole of Mg reacted Mole of Hz formed (based on the stoichiometric equation for the reaction) Vapor Pressure of water from curve in mmHg Pressure of H2 from Dalton's Law in mme Pressure of H2 in atm Volume of H2 in liter (Using volume measured in Equalization Chamber) Temperature in Kelvin Calculation of R (6 pts, each) Percent error Average R = 63Explanation / Answer
Balanced equation:
Mg(s) + 2 HCl(aq) = H2(g) + MgCl2(aq)
For trial 1
Moles of Mg reacted = mass/molecular weight
= 0.036g / 24.3 g/mol
= 0.00148 mol
Moles of H2 formed = moles of Mg consumed = 0.00148 mol
Vapor pressure of water at 20.5 °C = 18.1 mmHg
Pressure of H2 = barometric - vapor pressure
= 731.2 - 18.1
= 713.1 mmHg
Pressure of H2 in atm = 713.1 mmHg x 1atm/760mmHg
= 0.9383 atm
Volume of H2 gas in L = 39.5 mL x 1L/1000 mL
= 0.0395 L
Temperature in K = 20.5 + 273.15 = 293.65 K
From the ideal gas equation
R = PV/nT
= 0.9383atm x 0.0395 L / 0.00148 mol x 293.65 K
= 0.08528 L-atm/mol-K
% error = (0.08528 - 0.08206)*100/0.08206
= 3.92%
For trial 2
Moles of Mg reacted = mass/molecular weight
= 0.037g / 24.3 g/mol
= 0.00152 mol
Moles of H2 formed = moles of Mg consumed = 0.00152 mol
Vapor pressure of water at 20.5 °C = 18.1 mmHg
Pressure of H2 = barometric - vapor pressure
= 731.2 - 18.1
= 713.1 mmHg
Pressure of H2 in atm = 713.1 mmHg x 1atm/760mmHg
= 0.9383 atm
Volume of H2 gas in L = 39.5 mL x 1L/1000 mL
= 0.0395 L
Temperature in K = 20.5 + 273.15 = 293.65 K
From the ideal gas equation
R = PV/nT
= 0.9383atm x 0.0395 L / 0.00152 mol x 293.65 K
= 0.08304 L-atm/mol-K
% error = (0.08304 - 0.08206)*100/0.08206
= 1.19%
Average R = (0.08528 + 0.08304) /2
= 0.08416 L-atm/mol-K