In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

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

i believe it is 40–40–20%

explanation:

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2