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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.

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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
(2.0)/(2)* 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* \text{ Molar mass of }O_2


\text{ Mass of }O_2=(3.0moles)* (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=(w)/(M)RT\\\\P=(w)/(V)* (RT)/(M)\\\\P=\rho* (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* ((0.0821L.atm/mole.K)* (487K))/(32g/mol)


P=1.78atm

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

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