What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
2c2H6(g)+7O2(g)--->4CO2(g)+6H2O( g)

For the production of 77.4 L water 90.3 L oxygen is required.

Explanation:

Given data:

Volume of oxygen required = ?

Volume of water produced = 77.4 L

Solution:

Chemical reaction equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

1 mole = 22.414 L

There are 6 moles of water = 6×22.414 = 134.5 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

Now we will compare the litters of water and oxygen:

                              H₂O           :              O₂    

                              134.5         :              156.9

                                77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol