(Unless otherwise noted, 2 pts. for each blank) Molarity of HCI (M) (1pt, each):
ID: 1038705 • Letter: #
Question
(Unless otherwise noted, 2 pts. for each blank) Molarity of HCI (M) (1pt, each): Volume of HCI (mL): Mass of Mg (gram): Volume of gas before placing in Equalization Chamber (mL) (1 pt.) Volume of gas after placing in Equalization Chamber (mL) Barometric Pressure (mm Hg): Temperature C) Write the balanced overall equation for the reaction of magnesium and HCI to form hydrogen Data, Trial 1 Data, Trial 2 9. St2 4G S 19.a 045 S0ml (This value NOT used in cakculations below) 155mm H COmnd gaR0c gas and the magnesium salt (2 pts): 1.8 Mole of Mg reacted Mole of H2 formed (based on the stoichiometric equation for the reaction) 0 at OOm to Vapor Pressure of water from curve in mmg: Pressure of H2 from Dalton's Law in mmg: Pressure of H2 in atm Volume of H2 in liter Using volume measured in Equalization Chamber) Temperature in Kelvin 095 85k 395.95k Calculation of R (6 pts. each) Percent error Average R- 63Explanation / Answer
1) According to Dalton's law of partial pressure:
P(atm)=P(H2)+p(water vapour)
trial 1) p(H2)=P(atm)-p(water vapour)=755 mmhg-20.8 mmhg=734.2 mmhg
trial 2) p(H2)=P(atm)-p(water vapour)=760 mmhg-21.0 mmhg=739.0 mmhg
trial 1) P(atm)=734.2 mmhg*(1atm/760 mmhg)=0.966 atm
trial 2) P(atm)=739 mmhg*(1atm/760 mmhg)=0.972 atm
trial 1) volume of H2 in L=48.5 ml=0.0485 L
trial 2) volume of H2 in L=52.0 ml=0.052L
calculation of R,
using ideal gas eqn:
PV=nRT,where P=pressure of gas,V=volume of gas,T=temperature,R=ideal gas constant,n=mol of H2
trial 1) R=PV/nT =0.966 atm *(0.0485 L)/(1.72 mol)(295.85K)=9.207*10^-5 Latm/Kmol
trial 2) R=PV/nT =0.972 atm *(0.052 L)/(1.85 mol)(295.95K)=9.232*10^-5 Latm/Kmol
percent error=(actual value-experimental value)/(actual value)*100=(0.0821-9.207*10^-5)/(0.0821)*100=99.89%(trial 1)
trial 2: percent error=(0.0821-9.232*10^-5)/(0.0821)*100=99.89%
Av R=(9.207*10^-5)+(9.232*10^-5)/2=9.219*10^-5 Latm/Kmol
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