This is an incomplete question, here is a complete question.
Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.
Volume of rigid steel container: 1.00 L
Molar mass of Nitroglycerine: 227 g/mol
Temperature: 300 K
Amount of Nitroglycerine tested: 227 g
Value for ideal gas constant, R: 0.0821 L.atm/mol.K
In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?
Answer : The partial pressure of the water vapor is, 20.01 atm
Explanation :
First we have to calculate the moles of 

Now we have to calculate the moles of 
The balanced chemical reaction is:

From the balanced chemical reaction we conclude that,
As, 4 moles of
react to give 12 moles of 
So, 1 moles of
react to give
moles of 
and,
As, 4 moles of
react to give 1 moles of 
So, 1 moles of
react to give
moles of 
and,
As, 4 moles of
react to give 6 moles of 
So, 1 moles of
react to give
moles of 
and,
As, 4 moles of
react to give 10 moles of 
So, 1 moles of
react to give
moles of 
Now we have to calculate the mole fraction of water.


Now we have to calculate the partial pressure of the water vapor.
According to the Raoult's law,

where,
= partial pressure of water vapor gas = ?
= total pressure of gas = 58 atm
= mole fraction of water vapor gas = 0.345
Now put all the given values in the above formula, we get:


Therefore, the partial pressure of the water vapor is, 20.01 atm