A gas in a container has a pressure of 0.77 atm, volume of 65,5 L and a temperature of 235 K. If the gas is helium, what mass of helium is in the container?
A) 2.61 g
B) 10.5 g
C) 0.625 g
D) 0.859 g

B) 10.5

Explanation:

PV= NRT     -> N=PV/RT

N= mols .     P= pressure . V= volume in Liters    T= temp in Kelvin (Celsius degree = 273.15) .  R= Gas constant (.0821 for atm)

N=(.77 x 65.5)/(.0821x235) = 2.614 mols of Helium

Mols x Molecular weight = weight in grams

2.614x4.003(Molecular weight ot He) = 10.46 round to 10.5  

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :