Limiting reagent problem? For the reaction #C_2H_6+O_2 to CO_2+H_2O#

Balanced Eqn
#2C_2H_6 +7O_2 =4 CO_2+6H_2O#
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume #60*300/224 = 80.36 g# ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced =#4*44*80.36/60 =235.72g#

and its no. of moles will be #235.72/44# =5.36 where 44 is the molar mass of Carbon dioxide

Balanced equation is #C_2H_6+7/2O_2 to 2CO_2+3H_2O#

Alternatively, this can be written as #2C_2H_6+7O_2 to 4CO_2+6H_2O#

Find the number of moles of each reactant:

For #C_2H_6#, with molar mass #30# #gmol^-1#:

#n=m/M=270/30=9# #mol#

For #O_2#, with molar mass #32# #gmol^-1#:

#n=m/M=300/32=9.375# #mol#

It might look as though we have an excess of #O_2#, but remember that each #1# #mol# of #C_2H_6# required #7/2# #mol# of #O_2#, so in fact we have less oxygen than we need. Some #C_2H_6# will remain unburned at the end of the reaction.

Divide the #9.375# #mol# by #7/2# to discover that #2.68# #mol# of #C_2H_6# will react. Each mole of #C_2H_6# produces #2# #mol# of #CO_2#, so #5.36# #mol# of #CO_2# will be produced.

The remaining amount of #C_2H_6# will be #9-2.68=6.32# #mol#. To find the mass of excess #C_2H_6#:

#m=nM=6.32*30=189.6# #g#