Question

How many grams of KCl (s) has been produced from the thermal decomposition of KClO3 (s) that produced 50.0 mL of O2 (g) at 25 degrees C and 1.00 atm pressure.

Answer 1

The mass is
`m_(KCl) = 0.102 \ g`

Step-by-step explanation:

From the question we are told that

The volume of oxygen produced is
`V_o = 50mL = 50 *10^(-3) L`

The temperature is
`T = 25 ^oC = 25* 273 = 298 K`

The pressure is
`P = 1.0\ atm`

From the ideal gas law we have that

`PV = nRT`

Where R is the gas constant with the value

`R = 0.08206 \ atm \cdot L /mol K`

n is the number of moles making it the subject of the formula

`n = (PV)/(RT)`

Substituting values

`n = (1 * (50 *10^(-3)) )/(0.08206 * 298)`

`n = 2.045 *10^(-3) \ mol`

From the chemical equation

one mole of
`KClO_3` produces one mole of kCl and
`(3)/(2)` of oxygen

x mole of
`KClO_3` produces x mole of kCl and
`2.045 *10^(-3) \ mol` of oxygen

So
`x = (2.045 *10^(-3))/((3)/(2) )`

`x = (2)/(3) * 2.045 *10^(-3)`

`x = 1.363 *10^(-3) \ mol`

Now the molar mass of KCl is a constant with a value

`M_(KCl) = 74.55 g/mol`

Now the mass of KCl is mathematically evaluated as

`m_(KCl) = x * M_(KCl)`

Substituting values

`m_(KCl) = 1.363 *10^(-3) * 74.55`

`m_(KCl) = 0.102 \ g`