Section/Group Instructor: Post-Lab Report (Use the In-lab observations to comple

Question

Section/Group Instructor: Post-Lab Report (Use the In-lab observations to complete the laboratory report. Turn in to your Instructor when you have completed the report.) mm H20 Sample 2 Sample ID of sample Buret reading of vol. of H, collected mL mL vol. in 43.5 mL mL of H2 graduated portion of buret Volume of buret below 50-mL mark mL mL mL Total volume of H, collected 755 torr torr Barometric pressure Difference in water levels 4,8 torr torr Correction for difference in water levels torr torr Pressure of gas in buret Temperature of water bath Water vapor pressure at water bath temp. torr torr torr Pressure of dry hydrogen mL Volume H2 corrected to STP mol mol Moles H2 collected Mass Mg in sample Show your work for the determination of the volume of gas collected for sample 1. Include calculation of volume inside the graduated portion of the buret from buret reading and use of void volume. Experiment 13 Determination of the Mass of Magnesium Metal in a sample 19

Explanation / Answer

Volume of hydrogen collected ( trial-1)V2 = 43.5 ml=43.5/1000L= 0.0435 L, Pressure,P2=742.93 Torr= 742.93/760 atm=0.9775 atm, Temperature, T2 = 23 deg,c= 23+273= 296K

From gas law standard conditions refer to P1=1atm, T1= 273 K,

From P1V1/T1= P2V2/T2, V1= P2V2*T1/(T2*P1)= 0.9775*0.0435*273/(296*1)= 0.039L

22.4 Liters are there in 1 mole of any gas

0.039L will be there in 0.039/22.4= .00174

The reaction of Mg and HCl to generate H2 is Mg+ 2HCl ----àMgCl2+ H2

1 mole of H2 requires 1 mole of Mg

0.00174 moles of H2 required 0.00174 moles of Mg.

Atomic   mass of Mg =24, mass of Mg= Moles* atomic mass=0.00174*24=0.042 gm

Trial-2

Volume of hydrogen collected V2 = 50.4 ml=50.4/1000L= 0.0504 L, Pressure,P2=742.93 Torr= 742.93/760 atm=0.9775 atm, Temperature, T2 = 23 deg,c= 23+273= 296K

From gas law standard conditions refer to P1=1atm, T1= 273 K,

From P1V1/T1= P2V2/T2, V1= P2V2*T1/(T2*P1)= 0.9775*0.0504*273/(296*1)= 0.045L

22.4 Liters are there in 1 mole of any gas

0.045L will be there in 0.045/22.4= .002

The reaction of Mg and HCl to generate H2 is Mg+ 2HCl ----àMgCl2+ H2

1 mole of H2 requires 1 mole of Mg

0.00174 moles of H2 required 0.002moles of Mg.

Atomic   mass of Mg =24, mass of Mg= Moles* atomic mass=0.002*24=0.048 gm

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