Respuesta :
Answer:
Moles of potassium chlorate = 0.02976 moles
Explanation:
At standard pressure and temperature,
22.4 L of a gas consists of 1 mole
Thus, given, volume of [tex]O_2[/tex] = 1.0 L
So,
1 L of a gas consists of [tex]\frac{1}{22.4}[/tex] mole
Moles of oxygen gas = 0.04464 moles
The reaction is shown below as:-
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.
So,
1 mole of oxygen gas are produced when [tex]\frac{2}{3}[/tex] moles of potassium chlorate undergoes reaction.
Thus,
0.04464 mole of oxygen gas are produced when [tex]\frac{2}{3}\times 0.04464[/tex] moles of potassium chlorate undergoes reaction.
Moles of potassium chlorate = 0.02976 moles
From the decomposition reaction 2KClO₃(s) → 2KCl(s) + 3O₂(g), the number of moles of KClO₃ to be decomposed to generate 1.0 L of O₂ gas at standard temperature and pressure (STP) is 0.030.
The balanced chemical reaction for the decomposition of potassium chlorate (KClO₃) is the following:
2KClO₃(s) → 2KCl(s) + 3O₂(g) (1)
We can find the number of moles of O₂ gas with the Ideal gas equation:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 1.0 atm (at STP conditions)
V: is the volume = 1.0 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 273 K (at STP conditions)
n: is the number of moles =?
The number of moles of O₂ gas is:
[tex] n_{O_{2}} = \frac{PV}{RT} = \frac{1.0 atm*1.0 L}{0.082 L*atm/(K*mol)*273 K} = 0.045 \:moles [/tex]
From reaction (1), we have that 2 moles of KClO₃ produce 3 moles of O₂, so the number of moles of KClO₃ resulting from the decomposition is:
[tex] n_{KClO_{3}} = \frac{2\:moles\:KClO_{3}}{3\:moles\:O_{2}}*0.045\:moles\:O_{2} = 0.030 \:moles [/tex]
Therefore, the number of KClO₃ moles to be decomposed is 0.030.
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