Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.
Explanation : Given,
Moles of
= 2.0 moles
Molar mass of
= 32 g/mole
Now we have to calculate the moles of ![MgO](https://tex.z-dn.net/?f=MgO)
The balanced chemical reaction is,
![2KClO_3\rightarrow 2KCl+3O_2](https://tex.z-dn.net/?f=2KClO_3%5Crightarrow%202KCl%2B3O_2)
From the balanced reaction we conclude that
As, 2 mole of
react to give 3 mole of ![O_2](https://tex.z-dn.net/?f=O_2)
So, 2.0 moles of
react to give
moles of ![O_2](https://tex.z-dn.net/?f=O_2)
Now we have to calculate the mass of ![O_2](https://tex.z-dn.net/?f=O_2)
![\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DO_2%3D%5Ctext%7B%20Moles%20of%20%7DO_2%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DO_2)
![\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DO_2%3D%283.0moles%29%5Ctimes%20%2832g%2Fmole%29%3D96g)
Therefore, the mass of oxygen gas produced is, 96 grams.
Now we have to determine the pressure exerted by the gas against the container walls.
Using ideal gas equation:
![PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5CPV%3D%5Cfrac%7Bw%7D%7BM%7DRT%5C%5C%5C%5CP%3D%5Cfrac%7Bw%7D%7BV%7D%5Ctimes%20%5Cfrac%7BRT%7D%7BM%7D%5C%5C%5C%5CP%3D%5Crho%5Ctimes%20%5Cfrac%7BRT%7D%7BM%7D)
where,
P = pressure of oxygen gas = ?
V = volume of oxygen gas
T = temperature of oxygen gas = ![214.0^oC=273+214.0=487K](https://tex.z-dn.net/?f=214.0%5EoC%3D273%2B214.0%3D487K)
R = gas constant = 0.0821 L.atm/mole.K
w = mass of oxygen gas
= density of oxygen gas = 1.429 g/L
M = molar mass of oxygen gas = 32 g/mole
Now put all the given values in the ideal gas equation, we get:
![P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}](https://tex.z-dn.net/?f=P%3D1.429g%2FL%5Ctimes%20%5Cfrac%7B%280.0821L.atm%2Fmole.K%29%5Ctimes%20%28487K%29%7D%7B32g%2Fmol%7D)
![P=1.78atm](https://tex.z-dn.net/?f=P%3D1.78atm)
Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.