Question

1. Sulfur hexafluoride, SF6, is a gas that a potent greenhouse gas and electrical insulator. If a 265.0 g

sample is collected at 825.3 torr and at 13.7°C, what volume does this sample occupy?

2. A sample of an organic compound weighs 0.495 g. This sample is a gas collected in a flask that has

an exact volume of 127 mL. At 98°C, the pressure of the gas is 754 mmHg in the flask. Determine

the number of moles for the compound, and use that to calculate the molecular weight of the organic

compound in the flask.

3. Octane burns in air according to the following unbalanced equation:

C8H18(g) + O2(g) ® CO2(g) + H2O(g)

a. If 75.0 g of octane react with excess oxygen, how many moles of carbon dioxide will be

generated?

b. How many liters of carbon dioxide will be generated if it is collected at 0.985 atm and 23.0°C?

4. What is the density of UF6 at 100°C and 1 atm? It is a very dense gas.

5. Elemental sulfur can be recovered from gaseous hydrogen sulfide through the reaction:

2 H2S (g) + SO2 (g) ® 3 S (s) + 2 H2O (l)

What volume of H2S (in L, at 0.00°C and one atm) is required to produce 2.00 kg of sulfur by this

process?

6. Sulfur dioxide reacts with oxygen to give sulfur trioxide. Suppose that at one stage of the reaction

26.0 mol SO2, 83.0 mol O2, and 17.0 mol SO3 are present in the reaction vessel, and the total

pressure of the system is 0.950 atm. Calculate the mole fraction of SO3 and its partial pressure.

7. The following arrangement of flasks is set up. Assuming no temperature change and no reaction,

determine (a) the final pressure inside the system after all stopcocks are opened and (b) the mole

fraction of each gas. Assume that the connecting tube has no volume.

l________________________l______________________________l

O₂(g)                                    N₂(g)                                                    Ar (g)

V= 5.00 L                             V= 4.00 L                                             V=3.00 L

P= 2.51 atm                         P=0.792 atm                                        P= 1.23atm

Answer 1

Weight of SF6

.Molar mass of SF6 is 146.0554 g/mol

Moles of SF6 = 256 / 146.0554 = 1.7527 Moles

PV = nRT

P = 825.3 Torr = 1.0859211 atm

V= ?

n = 1.7527 Moles

R = 0.0821 L atm K-1 Mol-1

T = 273 + 13.7 = 286.7 K

Use the values in the formula

1.0859211 atm x V L = 1.7527 mol x 0.0821 L atm K-1 Mol-1 x 286.7 K  

V = 37.99 Liters ~ 40 Liters

Question 2

PV = nRT

P = 754 mm Hg = 0.9921 atm

V= 127 ml = 0.127 Liter

n = ?

R = 0.0821 L atm K-1 Mol-1

T = 273 + 98 = 371 K

0.9921 x 0.127 = n x 0.0821 x 373

n = 0.0041 Moles

Molecular weight of the compound = 0.495 g / 0.0041 mol = 119.66 g /mol

Question 3

Balanced equation:
2 C₈H₁₈(g) + 25 O₂(g) = 16 CO₂(g) + 18 H₂O(g)

75 gm of Octane = 75 / 114.22 =  0.6565 Moles

moles of carbon dioxide will be generated = 0.6565 x 16 / 2 =  5.2526 Moles

Volume of CO2  

P = 0.985 atm

V = ?

n = 5.2526 Liters

R = 0.0821 L atm K-1 Mol-1

T = 296 K

0.985 x V = 5.5226 x 0.0821 x 296

V = 129.59 Liter of CO2 will be generated