Question

Hi! I really need help:)

a) What is the average molar mass of a mixture of gases if 34.6g occupy 45.6L at 57.9 degrees C and 1045 mmHg?

b) Determine the liters of Oxygen needed for the complete combustion of 769g of benzene (C6H6) if the teaction is run at 23.5 degrees C and 765mHg?

42. What is the average molar mass of a 57.9°C and 1045mmHg? af mass of a mixture of pases if 34.6g occupy 45.0L 43. Determine the liters of Oxygen need for the complete combustion of 768g of benzene (C.H.) if the reaction is run at 23.5°C and 765mHg? C.Han + Op → _CO3 + __H2018)

Answer 1

From ideal gas equation, PV = nRT

PV = (mass of the mixture*RT)/ average molar mass of a mixture

average molar mass of a mixture = mass of the mixture * RT/PV

average molar mass of a mixture = (36.4*0.0821*330.9)/(45.6*1045/760) = 15.77 gms

b.

The number of moles of benzene = weight /MW = 768/78 = 9.85 MOLES

The balanced chemical reaction for the combustion of benzene is

2C6H6(l)    + 15 O2(g)    -------------> 12CO2(g) + 6H2O(g)

From the balanced equation, 2 moles of benzene requires 15 moles of OXYGEN gas

9.85 moles of benzene require 9.85*15/2 = 73.875 moles of oxygen.

From ideal gas equation, PV = nRT

volume of oxgen required for the complete combustion of 768 gms of benzene,

V = nRT/P = (73.875*0.0821*296.5)/(765/760) = 1786.6 liters