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

(1) Balance
(2) if "270 g" "C"_2"H"_6 and "300 g" "O"_2 reacted, how many moles of "CO"_2 were produced?
(3) How many grams excess leftover?

2 Answers
Feb 16, 2016

see below

Explanation:

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

Feb 18, 2016

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

Explanation:

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