Question

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?

Answer 1

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

Answer 2

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`