Answer:
The mass is [tex]m_{KCl} = 0.102 \ g[/tex]
Explanation:
From the question we are told that
The volume of oxygen produced is [tex]V_o = 50mL = 50 *10^{-3} L[/tex]
The temperature is [tex]T = 25 ^oC = 25* 273 = 298 K[/tex]
The pressure is [tex]P = 1.0\ atm[/tex]
From the ideal gas law we have that
[tex]PV = nRT[/tex]
Where R is the gas constant with the value
[tex]R = 0.08206 \ atm \cdot L /mol K[/tex]
n is the number of moles making it the subject of the formula
[tex]n = \frac{PV}{RT}[/tex]
Substituting values
[tex]n = \frac{1 * (50 *10^{-3}) }{0.08206 * 298}[/tex]
[tex]n = 2.045 *10^{-3} \ mol[/tex]
From the chemical equation
one mole of [tex]KClO_3[/tex] produces one mole of kCl and [tex]\frac{3}{2}[/tex] of oxygen
x mole of [tex]KClO_3[/tex] produces x mole of kCl and [tex]2.045 *10^{-3} \ mol[/tex] of oxygen
So [tex]x = \frac{2.045 *10^{-3}}{\frac{3}{2} }[/tex]
[tex]x = \frac{2}{3} * 2.045 *10^{-3}[/tex]
[tex]x = 1.363 *10^{-3} \ mol[/tex]
Now the molar mass of KCl is a constant with a value
[tex]M_{KCl} = 74.55 g/mol[/tex]
Now the mass of KCl is mathematically evaluated as
[tex]m_{KCl} = x * M_{KCl}[/tex]
Substituting values
[tex]m_{KCl} = 1.363 *10^{-3} * 74.55[/tex]
[tex]m_{KCl} = 0.102 \ g[/tex]
