Question

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

Answer 1

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:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

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 $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas  = ?

$p_T$ = total pressure of gas  = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas  = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

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

Answer 2

The partial pressure exerted by a gas in a mixture, depends on the mole

fraction of the gas and the pressure exerted by the mixture.

The partial pressure  of H₂O is 20 atm.

Reasons:

Given parameters are;

Explosion equation is 4C₃H₅N₃O₉ → 12CO₂(g) + O₂(g) + 6N₂(g) + 10H₂O(g)

Amount of nitroglycerine = 227 g

Molar mass of nitroglycerine = 227 g/mol

Required:

Partial pressure of the water vapor

Solution:

Number of moles of nitroglycerine in the reaction = 1 mole

Therefore;

Number of moles of CO₂ = 12/4 = 3 moles

Number of moles of O₂ = 0.25 moles

Number of nitrogen, N = 1.5 moles

Number of moles of H₂O = 2.5 moles

$Mole \ fraction \ of \ H_2O, \ X_{H_2O} = \dfrac{2.5}{2.5 + 1.5 + 0.25 + 3} = \dfrac{10}{29}$

According to Raoults law, we have;

The partial pressure of H₂O = $X_{H_2O} \times P_$

Therefore,  partial pressure  of H₂O = $\dfrac{10}{29} \times 58$ = 20 atm.

Learn more here:

https://brainly.com/question/10165688