CALCULATIONS: 1)Mass of Mg ? 2)Moles of Mg ? 3) Barometric pressure in torr.?4.

Question

CALCULATIONS:

1)Mass of Mg ?

2)Moles of Mg ?

3) Barometric pressure in torr.?4. Partial pressure of Hydrogen gas in torr and atmospheres.?5 Estimated temperature(room and water temperature before and after reaction) of the H2 gas in degrees C and then in Kelvin.

4) Moles of H2 gas (use PV=nRT) ?

5) Ratio of moles of Mg to moles H2. ?

6) The volume that the sample of H2 gas would occupy at STP (use: P1V1/T1 = P2V2/T2). ?

7)Using the volume from #8 and the moles of H2 (moles H2 = moles Mg) from #2, ?calculate the number of liters at STP per mole of H2 gas. ?

Info :

mass of Mg per meter : 1.841 g/m

lenght of Mg ribbon( to nearest 0.2 ml): 48.0 mm

room temp : 19.5 C

beaker water temp : (a) befor reaction : 19.5 C (b) after reaction : 21 C

volume of gas collected : 90.15 mL

barometic pressure : 75.16 cm Hg

vapor pressure of water 18.659 at 21 C

Explanation / Answer

1) Mass of Magnesium is found by equation,

linear density = mass (g) / length

We have her,

linear density = 1.841 g/m

length = 48 mm = 0.048 m

Substituting the values we get,

1.841 = mass (g) / 0.048

solving for mass (g) of magnesium = 1.841 x 0.048

                                                            = 0.0884 g

2) moles of magnesium is calculated using the relation,

moles = g / molar mass

molas mass of Mg = 24.305 g/mol

g as calculated from 1) = 0.0884 g

Substituting the values we get,

moles of Mg = 0.0884 / 24.305

                      = 0.00364 moles
3) We know that,

76 cm Hg = 760 torr

Given, barometric pressure = 75.16 cm Hg

Thus, pressure in torr of barometer = 75.16 x 760 / 76

                                                            = 751.6 torr

Partial pressure of H2 calculation.

P(H2) = P(barometer) - P(H2O)

Given here are,

Partial pressure of water = 18.659

Boromeric Pressure = 751.6 torr

Feeding the values we get,

P(H2) = 751.6 - 18.659

          = 732.941 torr

P(H2) in atm

we know 1 torr = 0.0013 atm

Thus, 732.941 torr = 0.964 atm is the pressure of H2

Estimated temperature of H2 gas

Before = 19.5 C = 19.5 + 273 = 292.5 K

After = 21 C = 21 + 273 = 294 K

4) moles of H2 gas by relation

PV = nRT

n = PV/RT

We have here,

P = 0.964 atm

T = 294 K

Volume = 90.15 mL = 0.09015 L

R = 0.0831 L.atm/mol.K

Substituting the values we get,

n = 0.964 x 0.09015 / 0.0831 x 294

   = 0.0036 moles

5) Ratio of moles of Mg to H2

ratio = 0.00364 / 0.0036

        = 1.0

6) Volume of H2 at STP

P1V1/T1 = P2V2/T2

Given are,

P1 = 0.9869 atm

P2 = 0.964 atm

V1 = unknown

V2 = 0.09015 L

T1 = 273.15 K

T2 = 294 K

Substituting the values we get,

0.9869 V1/273.15 = 0.964 x 0.09015 / 294

Solving for V1 = 0.08 L

Thus, volume of H2 gas at STP = 0.08 L

7) Number of Litres of H2 per L

We have,

volume = 0.08 L

moles of H2 = 0.00364 moles

Thus, number of L at STP per mol = 0.08 / 0.00364

                                                          = 22 L

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