Part A: Use molar volume to calculate each of the following at STP: the volume,

Question

Part A: Use molar volume to calculate each of the following at STP: the volume, in liters, occupied by 2.90 moles of N2 gas and the number of moles of CO2 in 3.20 L of CO2 gas.

Part B: A 10.0 g sample of krypton has a temperature of 25C at 574 mmHg . What is the volume, in milliliters, of the krypton gas?

Part C: A mixture of He, Ar, and Xe has a total pressure of 2.00 atm . The partial pressure of He is 0.450 atm , and the partial pressure of Ar is 0.300 atm . What is the partial pressure of Xe?

Part D: 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?

Explanation / Answer

1 mole of any gas occupied volume is 22.4L

1 mole of N2 gas occupied volume is 22.4L

2.9 moles of N2 gas occupied volume is = 22.4*2.9 = 64.96 L

22.4 L of Co2 gas = 1 mole

3.2 L of Co2 gas = 3.2/22.4 =0.14 moles of Co2

part-B

PV = nRT

no of moles n = W/G.A.Wt   = 10/83.8 = 0.119 moles

T = 25C0 = 25+273 = 298K

P = 574mmHg = 574/760 = 0.755atm

R = 0.0821L-atm/mole-K

V = nRT/P

   = 0.119*0.0821*298/0.755 = 3.85L

volume = 3850ml

part-c

PT   = PHe + PAr + PXe

PHe = 0.45atm

PAr = 0.3atm

PT = 2atm

2    = 0.45+ 0.3+ PXe

PXe = 2-0.75 =1.25 atm

Part-D

PV = nRT

n = nN2 + nO2 + nHe

V = 18L

T = 0C0 = 273K

P = 1atm

n= PV/RT

= 1*18/0.0821*273

= 18/22.4 = 0.8 moles

n = nN2 + nO2 + nHe

0.8 = 0.25 + 0.25 + nHe

nHe = 0.8-0.5 = 0.3 moles

mass of He = no of moles * atomic mass

                  = 0.3*4 = 1.2 gm >>>>> answer

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