Question

1. A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He . The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

2. A gas sample has a unknown pressure with a temperature of 51 ∘C . The same gas has a pressure of 2.96 atm when the temperature is -50. ∘C , with no change in the volume or amount of gas. What was the initial pressure, in atmospheres, of the gas?

2b.How many grams of NO2 will be produced when 2.1 L of nitrogen at 820 mmHg and 25 ∘C are completely reacted? 2c.What volume of O2 at 760. mmHg and 31 ∘C is required to synthesize 21.5 mol of NO ?

Answer 1

1]

PV = nRT

n = total moles = 0.25+0.25 + x [ ASsumed x moles of He ]

V = 18 L

T = 0 C = 273 K

R = 0.0821

P = 1 atm

1*18 = n *0.0821*273

n = 0.8

x = 0.8- (0.5) = 0.3

Moles of He = 0.3

1 mole = 4 gms

0.3 moles = 1.2 gms of He

2]

When volume is constant

PV = nRT ---> [ V , n, R are constants then ]

P1 / P2 = T1 / T2

x / 2.96 = 51+273 / -50+273

x = 4.3 atm

Initial pressure = 4.3 atm

2b ]

PV = n RT

P = 820/760 = 1.08 atm

V = 2.1 L

T = 25+273 = 298 K

(1.08)(2.1) = n (0.0821) (298)

n = 0.0927 mol N2

2. Calculate moles of NO2 formed:

N2 + 2O2 ---> 2NO2

0.0927 mol N2 X (2 mol NO2/1 mol N2) = 0.185 mol NO2

3. Calculate mass of NO2:

0.185 mol NO2 X 46.0 g/mol = 8.53 grams NO2

2c]

N2 + O2 ----> 2NO

2 moles of NO needs 1 mole of O2

to produce 21.5 mole of NO we need 21.5 / 2 = 10.75 moles of O2

PV = nRT

760 mm Hg = 1atm

1*V = 10.75*0.0821*(31+273)

V = 268.3 L