Answer the blank questions Blue pen: Trial 1 :Gas Volume not in Graduated Cylind
ID: 966392 • Letter: A
Question
Answer the blank questions
Blue pen: Trial 1 :Gas Volume not in Graduated Cylinder : 74.3 mL.
Blue pen: Trial 1 :Gas Volume not in Graduated Cylinder : 75.0 mL.
Chem 1A Molar Volume of a Ga Name Partner Molar Volume of a Gas Room temperature (T): 21 - Barometric pressure: 11214 tom torr Trial I Trial 2 Mass of Mg strip Temperature ofwater in the 600-ml beaker Vapor pressure of water at the temperature 2tc4c -te in the 600-mL beaker Gnas violume vnoo un ofgas collecvied at room tempera anas vd ne no olume ofgascollected at room temperatura 75mLt omLmo ml. nor in Grad cyl . m Volume of gas in liters (V:) 1. Partial pressure of hydrogen gas at room torr tom torr 2. Partial pressure of hydrogen gas in atm (P:) atmatnm atm 3. Volume of hydrogen gas at STP L. P (solve for VSTP) T: PSTPkm = 4. Moles of hydrogen produced based on amount mol mol mol of Mg used (use balanced equation) 5. Molar volume of hydrogen (Linol) at STP Lino! Linol | Umc1 (calculated from your data) Average molar volume of hydrogen Molar volume of an ideal gas at STP 6. Umol Lmol 7. Percent errorExplanation / Answer
Barometric pressure = 767.4 Torr
Partial pressure of hydrogen gas at room temperature= 767.4- 19.8 =747.6 Torr
Partial pressure of hydrogen gas in atm= 747.6/760 atm=0.983 atm ( trial-1 and trial-2)
Volume of gas= 75ml= 75/1000= 0.075 L at 22 deg.c= 22+273.15= 295.15K
From P1V1/T1= P2V2/T2
P1, V1 and T1 refer to standard condtions while P2, V2 and T2 refer to given conditions
1*V1/273.15= 0.983*0.075/295.15
V2= 0.983*0.075*273.15/295.15 =0.06823 L
For trial-2
V2= 0.983*0.077*273.15/295.15 = 0.07005 L
Mass of Mg strip= 0.074 g for both strips
Moles of Mg= 0.074/24= 0.003083 moles
Mg+2HCl-------à MgCl2+H2
As per the reaction, 1 mole of Mg produces one mole of hydrogen, moles of hydrogen= 0.003083 ( for both the trials)
Molar volume=0.06823/0.003083=22.13 L/mol ( Trial-1) at STP
Molar volume= 0.07005/0.003083=22.72 L/mol at STP ( Trial-2)
Average= (22.13+22.72)/2 L/mol=22.425 L/mol
Percentage error=100* (22.425-22.4)/22.4=0.11%
The possible error is measuremenet of volume which leads to error. Also measurment of mass of Mg also gives rise to error.
2. The reaction of sodlum nitrate i heating is
2NaNO3 (s) 2NaNO2 (s) + O2 (g)
Molecualr weight of NaNO3= 23+14+48= 85, moles of NaNO3 in 0.123 g= 0.123/85=0.001447
Molecular weight of NaNO2= 23+14+32=69
Molecular weight of oxygen =32
2 moles of NaNO3 gives 1 mole of oxygen
0.001447 moles of NaNO3 gives 0.001447/2=0.000724 moles of Oxygen
Theoretical mass of oxygen produces= 0.000724*32=0.023168 gms
Temperature of water= 20 deg.c, vapor pressure of water at 20 deg.c =17.5 mm Hg= 17.5 Torr
Partial pressure of Oxygen= 758-17.5=740.5 Torr= 740.5/760=0.974 atm
Volume of oxygen= 16.2ml =16.2/1000=0.0162L
Moles of oxygen ( Actual yiled )= PV/RT= 0.974*0.0162/(22.5+273.15)*0.08206=0.00065 moles =0.00065*32=0.020812 g,s
Percent yiled= 100*0.020812/0.023168= 89.83%