From the \"Determination of the Molar Volume of a Gas and the Gas Constant, R\"

Question

From the "Determination of the Molar Volume of a Gas and the Gas Constant, R" Lab:

(HCl, Magnesium Ribbon, Copper wire,etc. is used in the lab)

PLEASE DO ANSWER 1.e to Q2.

1. A student did the experiment described except she used Al foil instead of Mg. Her data table included the following values:

Mass of Al - 0.0352 grams

Temperature - 19.08 celsius

Barometric Pressure - 762.5 mmHg

Volume of H2 Gas - 46.92 mL

Answer the following questions based on this data table.

a. Write the balanced equation for the reaction between Aluminum and hydrochloric acid.

b. How many moles of Al were consumed?

c. How many moles of H2 were produced?

d. What was the vapor pressure of the water in the gas sample?

e. What was the pressure of the dry hydrogen?

f. Determine the experimental value of the gas constant, R.

g. Determine the volume of the dry H2 gas at STP conditions.

h. Determine the experimental value of volume of dry H2 gas at STP per mole of H2.

2. Why is there a lag time in this experiment between the inverting of the eudiometer tube and the start of the reaction between the metal and HCl?

Explanation / Answer

(a)

Balanced equation is,

2 Al (s) + 6 HCl (aq.) ------------> 2 AlCl3 (aq.) + 3 H2 (g)

(b)

Mass of Al = 0.0352 g.

Molar mass of Al = 27.0 g/mol

Moles of Al = mass / molar mass = 0.0352 / 27.0 = 0.00130 mol

(c)

From the balanced equation,

2 mol of Al produces 3 mol of H2

Then,

0.00130 mol of Al produces 3 * 0.00130 / 2 = 0.00196 mol of H2

(d)

Vapour pressure of water at STP condition = 23.75 mmHg

(e)

Vapour pressure of dry H2 gas = 762.5 - 23.75 = 738.75 mmHg = 738.75 / 760 = 0.972 atm

(f)

Ideal gas equation,

P V = n R T

0.972 * 0.04692 = 0.00196 * R * 273.15

R = 0.0852 L.atm.K-1.mol-1

(g)

At STP,

P = 1.00 atm and T = 273.15

1.00 * V = 0.00196 * 0.0852 * 273.15

V = 0.0456 L

(h)

Volume of dry H2 per mol = V / n = 0.0456 / 0.00196 = 23.3 L

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