Respuesta :
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 [tex]KCl_3[/tex] = 2.0 moles
Molar mass of [tex]O_2[/tex] = 32 g/mole
Now we have to calculate the moles of [tex]MgO[/tex]
The balanced chemical reaction is,
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]KClO_3[/tex] react to give 3 mole of [tex]O_2[/tex]
So, 2.0 moles of [tex]KClO_3[/tex] react to give [tex]\frac{2.0}{2}\times 3=3.0[/tex] moles of [tex]O_2[/tex]
Now we have to calculate the mass of [tex]O_2[/tex]
[tex]\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2[/tex]
[tex]\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g[/tex]
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:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}[/tex]
where,
P = pressure of oxygen gas = ?
V = volume of oxygen gas
T = temperature of oxygen gas = [tex]214.0^oC=273+214.0=487K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
w = mass of oxygen gas
[tex]\rho[/tex] = 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:
[tex]P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}[/tex]
[tex]P=1.78atm[/tex]
Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.
