Answer to Question #157825 in Physical Chemistry for sally

Question

When methane (CH4) is burned in oxygen, carbon dioxide and water vapor are formed.

When 8.00 L of methane at 125°C and 1.00 atm is burned in an excess of oxygen, the

products are collected into a separate 10.0 L flask.

(a) Write a balanced equation for the reaction.

(b) What is the total pressure of the products in the flask at 125°C?

(c) What is the partial pressure of each of the products in the flask?

Answer 1

(a).CH₄(g)+2O₂₍g)$\to$ CO₂(g) +2H₂O(g)

P₁V₁=P₂V₂

8*1=10P₂

(b).P₂=0.8atm

n=PV/RT

=1*8/(398*0.0820574)

=0.24496moles

Moles of CO₂=0.24496moles

Moles of H₂O=2*0.24496moles=0.48991moles

(c).P_CO2=0.24496*0.8/(0.48991+0.24496)

=0.2667

P_H2O=0.48991*0.8/0.734873

=0.5333