ALGÚN QUÍMICO!!?? Me podrían ayudar con este problema?:
Calcular la cantidad de clorato de potasio que será necesario descomponer para obtener 1 Kg de Oxígeno. ¿Qué volumen ocupará este en condiciones normales? (2KClO3 ------> 2KCl + 3O2).
Respuestas
Respuesta dada por:
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Mm KCl = 74.5 g/mol
KClO3 = 122.5 g/mol
mol KCl = 1000 g KCl x 1 mol KCl
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74.5 g
mol = 13.42 de KCl
g KClO3 = 13.42 mol KCl x 2 mol KClO3 x 122.5 g KCLO3
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2 mol KCl ` 1 mol KClO3
g de KClO3 = 1643.95 g
mol O2 = 13.42 mol KCl x 3 mol O2
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2 mol KCl
mol O2 = 20.13
V x P = n x R x T CN ( P = 1 atm y T = 273 K
R = 0.0821 (L atm/ mol K).
V = 20.13 mol x 0.0821 (L atm/ mol K).x 273`K
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1 atm
V = 451.18 LITROS
KClO3 = 122.5 g/mol
mol KCl = 1000 g KCl x 1 mol KCl
``````````````
74.5 g
mol = 13.42 de KCl
g KClO3 = 13.42 mol KCl x 2 mol KClO3 x 122.5 g KCLO3
`````````````````` ````````````````````
2 mol KCl ` 1 mol KClO3
g de KClO3 = 1643.95 g
mol O2 = 13.42 mol KCl x 3 mol O2
``````````````
2 mol KCl
mol O2 = 20.13
V x P = n x R x T CN ( P = 1 atm y T = 273 K
R = 0.0821 (L atm/ mol K).
V = 20.13 mol x 0.0821 (L atm/ mol K).x 273`K
`````````````````````````````````````````````````````` ``
1 atm
V = 451.18 LITROS
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