Question

If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls.

Answer 1

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 $KCl_3$ = 2.0 moles

Molar mass of $O_2$ = 32 g/mole

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 2.0 moles of $KClO_3$ react to give $\frac{2.0}{2}\times 3=3.0$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g$

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:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}$

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = $214.0^oC=273+214.0=487K$

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

$\rho$ = 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:

$P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}$

$P=1.78atm$

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.