Question

write the balanced equation for the complete combustion of ethene (c 2 h 4 ). if 2.70 mol c 2 h 4 are reacted with 6.30 mol o 2 , identify the limiting reactant. then, calculate the number of moles of water produced.

Answer 1

Moles of
`O_2` are 6.30,so
`O_2` is a limiting reactant & Total number of moles of water is 4.20.

Step-by-step explanation:

Given

For Ethene
`C_2H_4`

The Balanced Equation for the Combustion Reaction Of ethene is as follows :

`C_(2)H_4 + 3O_2 -- > 2CO_2 + 2H_2O`

now, Given that 2.70 moles
`C_2H_4` are reacted with 6.30 moles of
`O_2`

From the Balanced equation of combustion, We see that from the stoichiometry of reactants, 1 moles of
`C_2H_4` require 3 moles of
`O_2`

Hence ,

2.70 moles of
`C_2H_4` require (2.70*3= 8.10) moles of
`O_2`

So, clearly moles of
`O_2` are 6.30 ,so
`O_2` is limiting reactant , it limits the extent of reaction.

now,

Since
`O_2` is limiting reactant, so it will get fully consumed during the course of reaction

Hence 6.30 moles consume (6.30/3)= 2.10 moles of
`C_2H_4` (from stoichiometry)

Hence,

Total number of moles of water for 2.10 moles of
`C_2H_4` from stoichiometry

Water moles = 2*2.10 = 4.20 moles