Question

What mass (in g) of oxygen must be added to 9.64 g of neon at 23.3^oC, and in a 134 L contain for a final pressure of 907 mmHg?

Answer 1

Initially, we had 9.64 g of Neon

so, moles of Na = weight/molar mass = 9.64g/20.18g/mole

= 0.4778 moles

at temperature T = 23.3 C = 273+23.3 K = 296.3 K

and volume V = 134 L

By applying ideal gas equation PV=nRT where P is the pressure exerted by gas and R is the gas constant R = 0.0821 L atm /K-mole we have

Let the pressure of Ne gas P atm so,

P*134 L = 0.4778moles*0.0821 L atm/K-mole * 296.3 K

so, P = 0.4778*0.0821*296.3/134 = 0.08674 atm

P = 0.08674 atm = 0.08674*760 mmHg = 65.922 mmHg (1 atm = 760 mmHg)

Let the moles of O2 needs to be added to increase this pressure to 907 mmHg = x moles

so, the pressure of O2 gas must be 907mmHg - 65.922 mmHg = 841.078 mmHg

Pressure in atm = 841.078mmHg/760 mmHg = 1.1066 atm

Applying ideal gas equation

PV = xRT

1.1066atm*134L = x*0.0821Latm/k-mole*296.3 K

x = 1.1066*134/(0.0821*296.3) = 6.0956 moles

weight of 6.0956 moles of O2 gas = molar mass*moles

= 32g/mole*6.0956moles = 195.059 g

So, we need to add 195.059 g of oxygen.