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)](https://tex.z-dn.net/?f=4C_3H_5N_3O_9%28l%29%5Crightarrow%2012CO_2%28g%29%2BO_2%28g%29%2B6N_2%28g%29%2B10H_2O%28g%29)
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](https://tex.z-dn.net/?f=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](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DC_3H_5N_3O_9%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DC_3H_5N_3O_9%7D%7B%5Ctext%7BMolar%20mass%20%7DC_3H_5N_3O_9%7D%3D%5Cfrac%7B227g%7D%7B227g%2Fmol%7D%3D1mol)
Now we have to calculate the moles of ![CO_2,O_2,N_2\text{ and }H_2O](https://tex.z-dn.net/?f=CO_2%2CO_2%2CN_2%5Ctext%7B%20and%20%7DH_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)](https://tex.z-dn.net/?f=4C_3H_5N_3O_9%28l%29%5Crightarrow%2012CO_2%28g%29%2BO_2%28g%29%2B6N_2%28g%29%2B10H_2O%28g%29)
From the balanced chemical reaction we conclude that,
As, 4 moles of
react to give 12 moles of ![CO_2](https://tex.z-dn.net/?f=CO_2)
So, 1 moles of
react to give
moles of ![CO_2](https://tex.z-dn.net/?f=CO_2)
and,
As, 4 moles of
react to give 1 moles of ![O_2](https://tex.z-dn.net/?f=O_2)
So, 1 moles of
react to give
moles of ![O_2](https://tex.z-dn.net/?f=O_2)
and,
As, 4 moles of
react to give 6 moles of ![N_2](https://tex.z-dn.net/?f=N_2)
So, 1 moles of
react to give
moles of ![N_2](https://tex.z-dn.net/?f=N_2)
and,
As, 4 moles of
react to give 10 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
So, 1 moles of
react to give
moles of ![H_2O](https://tex.z-dn.net/?f=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}](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DH_2O%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2O%7D%7B%5Ctext%7BMoles%20of%20%7DH_2O%2B%5Ctext%7BMoles%20of%20%7DCO_2%2B%5Ctext%7BMoles%20of%20%7DO_2%2B%5Ctext%7BMoles%20of%20%7DN_2%7D)
![\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20%7DH_2O%3D%5Cfrac%7B2.5%7D%7B2.5%2B3%2B0.25%2B1.5%7D%3D0.345)
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](https://tex.z-dn.net/?f=p_%7BH_2O%7D%3DX_%7BH_2O%7D%5Ctimes%20p_T)
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:
![p_{H_2O}=X_{H_2O}\times p_T](https://tex.z-dn.net/?f=p_%7BH_2O%7D%3DX_%7BH_2O%7D%5Ctimes%20p_T)
![p_{H_2O}=0.345\times 58atm=20.01atm](https://tex.z-dn.net/?f=p_%7BH_2O%7D%3D0.345%5Ctimes%2058atm%3D20.01atm)
Therefore, the partial pressure of the water vapor is, 20.01 atm