3) A sample of nitrogen gas was initially at a pressure of 2.65 atm and a volume

Question

3) A sample of nitrogen gas was initially at a pressure of 2.65 atm and a volume of 515 mL. What volume will the gas occupy at a pressure of 0.90 atm?

3b) A 10.0 L tank is filled with oxygen under 10.0 atm and 25oC. Calculate the pressure in the tank after 10.0 g of oxygen is removed.

3c) To what value must the temperature be raised so that the pressure is maintained at 10.0 atm after theremoval of 10.0 g of oxygen?

3d) When 36 g of glucose ferments completely, how many grams of ethanol and carbon dioxide are produced? The fermentation reaction is C6H12O6 (aq) 2C2H5OH (aq) + 2CO2 (g).

3e) At 250C and under a pressure of 740 mm Hg, what is the volume of the carbon dioxide produced above?

Explanation / Answer

a)

Using the ideal gas equation

PV = nRT

Since the numbe of moles and Temperature remain the same, hence we can write

P1V1= P2V2

2.65 * 515 = 0.90 *V2

V2 = 1516.38 mL

3b)

Using the ideal gas equation

PV = nRT

(10)(10) = n * 0.0821 * (273+25)

n = 4.0873 moles

Number of moles of oxygen in 10g = 10/32 = 0.3125

n2 = n1 - 0.3125 = 4.0873 - 0.3125 = 3.7748 moles

Pressure = Pressure(1) * n2/n1 = 10 * 3.7748/4.0873 = 9.235 atm

3c)

n1T1 = n2T2

4.0873 * (273+25) = 3.7748 * T2

T2 = 322.67K = 49.67C

3d)

C6H12O6 ------- 2C2H5OH + 2CO2

Molar mass of glucose = 6 * 12 + 12 * 1 + 6 * 16 = 180 gm/mol

Number of moles = Mass/molar mass = 36/180 = 0.2 mol

Moles of ethanol produced = 0.4 moles

Molar mass of ethanol (C2H6O) = 2 * 12 + 6 * 1 + 16 = 46 gm/mol

Mass produced = 46 * 0.40 = 18.40 grams

Mass of CO2 produced = 36 - 18.40 = 17.60 grams

3e)

Using ideal gas equation

PV = nRT

740/760 * Volume = 0.40 * 0.0821 * (273+250)

Volume = 17.639L

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