Use the ideal gas law and the stoichiometry of a reaction lo dululllllllu magnes

Question

Use the ideal gas law and the stoichiometry of a reaction lo dululllllllu magnesium metal. MATERIALS Buret clamp and ring stand Buret 10 mL graduate cylinder Large beaker Magnesium metal Small beaker 6 M HCl solution Thermometer Barometer or other device for measuring atmospheric pressure Vapor Pressure of Water Near Room Temperature Vapor Pressure of H2O (torr) Temperature (C) 12.79 3.63 17 15.48 18 6.48 17.54 18.65 19.83 23 21.07 22.38 25 23.76 26 25.2 27 26.74 28 28.35 29 30.04 31.82 Advanced Chemistry with

Explanation / Answer

1.

Pressure of Dry H2 = total pressure – pressure of water vapor

= 76.2 mmHg – 19.83 mmHg

= 56.37 mmHg

= 0.0742 atm

2.

The ideal gas equation is as follows:

PV = n RT

n = PV/RT

here volume fro H2 gas = difference of water level = 37.25 mm

1 .00 mm= 10^-6 L

Then37.25 mm = 3.725 *10^-5 L

Temperature = 21.6 C = 294.75 K and R = gas constant

Therefore,

n = PV/RT

=0.0742 * 3.725 *10^-5 L / 0.08206 *294.75

= 1.15*10^-7 moles H2

3.

Mg(s) + 2 HCl (aq) -->MgCl2 + H2(g)

Now calculate the moles of Mg metal which are reacted in this reaction:

1.15*10^-7 moles H2 * 1 mole Mg/ 1.00 moles H2

1.15*10^-7 moles Mg

Amount of Mg in g =1.15*10^-7 moles Mg * molar mass,. 24.305 g/ mole

= 2.8*10^-6 g Mg

4.

Mg(s) + 2 HCl (aq) -->MgCl2 + H2(g)

Now calculate the number of mole of Mg in 0.039 g

= 0.039 g/ 24.305 g/moles

= 1.6*10^-3 moles

Then moles of    H2 which are theoretical produced:

1.6*10^-3 moles Mg * 1 mole H2/ 1.00 moles Mg

1.6*10^-3 mole H2

Percentage yield = observed moles of H2 / theoretical mole of H2*100

= 1.15*10^-7 moles H2 /1.6*10^-3 mole H2 *100

= 0.0072 %

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