I need answers to all the questions on both page Molar Volume of Hydrogen Gas &

ID: 304136 • Letter: I

Question

I need answers to all the questions on both page

Molar Volume of Hydrogen Gas & the Combined Gas Law Background The molar volume of a gas, as the name implies, is the volume on a per mole basis. At STP it has a value of 22.4 L/mol. The Combined Gas Law can be used to calculate the molar volume of a gas from non-STP, conditions 1, to STP conditions 2, as follows: ?2 T: Equation 1. T1 Equation 2 P2T Since V2 is the volume at STP, dividing it by the moles of gas, provides the molar volume of a gas at STP Equation 3. molar volume of a gas @STP moles of gas The gas used here is hydrogen which can be produced via a single replacement reaction between magnesium and hydrochloric acid: Mg (s) + 2HCl (aq) ? MgCl2 (aq) + H2 (g) The use of a eudiometer, a long calibrated glass tube, allows for the collection and volume measurement of the gas: a stoichiometric relationship between magnesium and hydrogen provides the moles of hydrogen gas. The combined gas law requires the measurement of Pi. Viand Ti for the collected gas once it has had a chance to sit a while to be certain all hydrogen has left the water, the temperatures have stabilized. and the water vapor in the eudiometer has reached its equilibrium pressure.

Explanation / Answer

3) The atmospheric pressure is PT = 765 mmHg and the temperature is 23.5°C, i.e, T1 = (273 + 23.5) K = 296.5 K.

The vapor pressure of water 23.5°C is PH2O = 21.1 mmHg.

The total pressure of the gas is given as

PT = PH2 + PH2O

====> 765 mmHg = PH2 + (21.1 mmHg)

====> PH2 = (765 mmHg) – (21.1 mmHg) = 743.9 mmHg (ans).

4) I need to know the value of V1 to calculate V2 (you haven’t provided either V1 or the moles of hydrogen gas liberated).

5) The molar volume of a gas is the volume occupied by one mole of the gas at STP. The molar volume of a gas has the value of 22.4 L/mol at STP (ans).

Step 1:

The atmospheric pressure is PT = 759 mmHg and the temperature of the gas is 22.0°C, i.e, T1 = (273 + 22.0) K = 295 K.

The vapor pressure of water 22.0°C is PH2O = 19.8 mmHg.

The total pressure of the gas is given as

PT = PH2 + PH2O

====> 759 mmHg = PH2 + (19.8 mmHg)

====> PH2 = (759 mmHg) – (19.8 mmHg) = 739.2 mmHg (ans).

Step 2:

Let us call PH2 from above as P1 = 739.2 mmHg.

STP is defined as the standard conditions of temperature of pressure, i.e, 273 K and 1 atm = 760 mmHg. Next, set up the table.

Initial Condition

Final Condition

P1 = 739.2 mmHg

T1 = 295 K

V1 = 85.5 mL

P2 = 760 mmHg

T2 = 273 K

V2 = to be determined

Step 3:

Use the combined gas law.

P1V1/T1 = P2V2/T2

=====> V2 = P1V1T2/P2T1 = (739.2 mmHg)*(85.5 mL)*(273 K)/(760 mmHg)(295 K)

=====> V2 = 76.958 mL ? 76.9 mL = (76.9 mL)*(1 L/1000 mL) = 0.0769 L (ans).

Step 4:

We start with 0.080 g magnesium.

The gram molar mass of magnesium is 24.305 g/mol.

Moles of Mg corresponding to 0.080 g = (0.080 g)/(24.305 g/mol) = 0.00329 mole (ans).

Step 5:

As per the stoichiometric equation,

1 mole Mg = 1 mole H2.

Therefore,

0.00329 mole Mg = 0.00329 mole H2 (ans).

Step 6:

Molar volume of H2 = (volume of H2)/(moles of H2) = (0.0769 L)/(0.00329 mole) = 23.374 L/mol ? 23.4 L/mol (ans).

Step 7:

The accepted value of the molar volume is 22.4 L/mol.

Percent error = ?(accepted value) – (obtained value)?/(accepted value)*100 = ?(22.4 L/mol) – (23.4 L/mol)?/(22.4 L/mol)*100 = 4.464% ? 4.46% (ans).