Question

Mass of flask, stopper, and 5mL of water (g)	102.945	g
Mass of flask, stopper, and filled with water (g)	238.389	g
Volume of gas space in flask (the difference between the above two measurements covnerted to mL) (L)	.135444	L

Mg(s) + HCl(aq) Reaction

	Trail 1	Trial 2
Mass of Mg(g)	0.008	0.007
Initial Pressure (atm)	0.9974	0.9945
Max Pressure (atm)	0.9980	0.007
Pressure Change (atm)	0.0006	0.0471
Temp (K)	295.7	296.1

a. Calculate the moles of H2 produced from each trial

b. Vol of gas was not at standard temp and pressure. Convert the volume at the pressure and temperature of the experiment to a volume at standard temperature and pressure. (STP = 273. 15K, P = 1.00atm)

c. Using your answers from part a and b, calculate the molar volume of H2(g) (L/mol) at STP for each trial.

(For what ever reason, I am only able to put in a good value for mols of H2 into the computer, it keeps saying my calculations are incorrect for the rest. Trying to figure out what I am doing wrong)

Answer 1

Mg(s) + HCl(aq) Reaction

2Mg + 2HCl = MgCl2 + H2

  PV= n RT

Here P = 0.0471atm

V= 0.135444 L

T= 295.7 K

n = PV/ RT

= 0.0006*0.135444 /0.08206* 295.7

= 3.35*10^-6 Moles

Trail 2:

2Mg + 2HCl = MgCl2 + H2

  PV= n RT

Here P = 0.0471atm

V= 0.135444 L

T= 296.1K

n = PV/ RT

= 0.0471*0.135444 /0.08206* 296.1

= 2.63*10^-4Moles

Vol of gas was not at standard temp and pressure. Convert the volume at the pressure and temperature of the experiment to a volume at standard temperature and pressure. (STP = 273. 15K, P = 1.00atm)

mole of any gas at STP occupies 22.4 liters of volume

thus, 3.35*10^-6 Moles will occupies = 22.4 L / 1 mol* 3.35*10^-6 Moles

= 7.504*10^-5L

Now calculate the molar volume

Volume ; 7.504*10^-5L / number of mole s

= 22.4 L / mole

And 2.63*10^-4 Moles will occupies = 22.4 L / 1 mol* 2.63*10^-4 Moles

= 5.89*10^-3 L

Now calculate the molar volume

Volume ; 5.89*10^-3 L / number of mole s2.63*10^-4

= 22.4 L / mole