Question

Mass of Magnesium strip Use the length of your magnesium strip and the mass per m provided by your TA to calculate the mass of your magnesium strip. Make sure to convert so that your units match! Mass of magnesium strip Moles of magnesium used Convert the mass of each magnesium strip into moles using the molar mass. Moles of magnesium Moles of hydrogen produced (n) Use the mole ratio from the balanced chemical equation to convert the moles magnesium into moles of Hz gas. Mg(s) + 2 HCl(aq) → Hz(g) + MgCl2(aq) Moles of hydrogen gas Pressure (P) You matched the level of water in the cylinder to that outside the cylinder so that you could assume that the gas in the cylinder was at atmospheric pressure. However, since you collected the gas over water, the pressure in the cylinder came from the vapor pressure of both the hydrogen and water. To accurately calculate R based on the hydrogen gas, you need to find the pressure exerted by the hydrogen gas only. Use Table 1 to find the vapor pressure of water at the temperature you measured in the lab and then use subtraction to calculate the pressure of H2 gas. Unless you saw a temperature change during the experiment, your pressure values should be the same for both trials. Make sure that all of your units are in atmospheres - conversions may be necessary! The most commonly used conversions are: 1 atm = 101.325 kPa = 760mm Hg = 760 torr The equation for the partial pressure of hydrogen is: Phydrogen = Patmosphere - Pwater Pressure of hydrogen gas

Answer 1

	Trial 1	Trial 2
Mass of magnesium (Mg) strip (g)	7.8 x 10-3 g	5.85 x 10-3 g
Moles of magnesium (mol)	3.21 x 10-4 mol	2.41 x 10-4 mol
Moles of hydrogen (H2) gas (mol)	3.21 x 10-4 mol	2.41 x 10-4 mol
Water Temperature (oC)	20.6 oC	20.6 oC
Vapour pressure (Pwater ) (atm)	0.024 atm	0.024 atm
Atmospheric pressure (Patmosphere ) (atm)	1 atm	1 atm
Pressure of hydrogen gas (Phydrogen = Patmosphere - Pwater ) (atm)	0.976 atm	0.976 atm
Volume of hydrogen gas (L)	0.0056 L	0.0043 L
Temperature of hydrogen gas (K)	293.75 K	293.75 K
Ideal gas constant (R) (L.atm.K-1.mol-1)	0.05796 L.atm.K-1.mol-1	0.05928 L.atm.K-1.mol-1
Average ideal gas constant	0.05862 L.atm.K-1.mol-1	-----------------------
Perecent error in R values	28.56 %	-----------------------

All calculations shown are for Trial-1. All calculations for Trial-2 are exactly similar so are not shown.

1. Length of the Mg ribbon = 0.4 cm = 0.004 m

Mass of 1.0 m Mg ribbon = 1.95 g

Mass of the 0.004 m Mg ribbon = (0.004 m)* (1.95 g/m) = 7.8 x 10-3 g

2. Moles of Mg = (mass of Mg taken)/ (molar mass of Mg)

= (7.8 x 10-3 g )/ (24.305 g/mol)

= 3.21 x 10-4 mol

3. Reaction: $Mg(s)+2HCl\;\rightarrow \;H_2(g)+MgCl_2 (aq)$

From the balanced reaction, 1 mole of Mg produces 1 mole of hydrogen.

Therefore, Moles of hydrogen, nhydrogen = moles of Mg = 3.21 x 10-4 mol

5. Vapour pressure at the temperature 20.6 oC, Pwater = 18.211 torr

$= 18.211\,torr\times \frac {1\,atm}{760\,torr} = 0.024\,\,atm$

6. Atmospheric pressure, Patmosphere = 1 atm

7. Phydrogen = Patmosphere - Pwater = (1 atm - 0.024 atm) = 0.976 atm

8. Volume of hydrogen gas, Vhydrogen = 5.6 mL = (5.6/ 1000) L = 0.0056 L

9. Temperature of hydrogen gas, Thydrogen = 20.6 oC = (20.6 + 273.15) K = 293.75 K

10. PhydrogenVhydrogen = nhydrogenRThydrogen

or, (subscripts have been dropped for simplicity)

R = PV nT (0.976 atm (0.0056 L) (3.21 x 10-4 mol (293.75 K)

or,

11. Average R

(0.05796) + (0.05928) Latm K-1 mol-1 = 0.05862 L atm K-1 mol-1

12. % error in R $= \frac {(actual\,value\,of\,\,R)-(experimental\,average\,value\,of\,\,R)}{actual\,value\,of\,\,R} \times 100$

  $= \frac {(0.08206\,L\,atm\,K^{-1}\,mol^{-1})-(0.05862\,L\,atm\,K^{-1}\,mol^{-1})}{0.08206\,L\,atm\,K^{-1}\,mol^{-1}} \times 100$

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