A #30*L# volume of dioxygen gas at #473*K#, and a pressure of #2*atm#, will completely combust WHAT mass of #"ethane gas"#?

1 Answer
Nov 14, 2015

Answer:

We need to work out (i) moles of oxygen gas, and (ii) equivalent moles of ethane.

Explanation:

#n=(PV)/(RT)# #=# #((2*atm)(30.0*L))/((0.0821*L*atm*K^(-1)*mol^-1)(473*K))#

#=# #1.54# #mol# #O_2(g)#.

Now, combustion reaction of ethane can be given as:
#C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(g)#

So, this #1.54# #mol# quantity represents 7/2 equiv of ethane by the stoichiometry of the rxn. Thus moles of ethane = #1.54*molxx2/7# #=# #0.440# #mol# ethane.

So (finally!), mass of ethane #=# #0.440*molxx30.07*g*mol^(-1)# #=# #??g#. Approx. 12 g?