cuando el aluminio reacsiona con el asido clorhidrico se produse cloruro de aluminio y gas hidrogeno calcula cuantos litros de hidrogeno medidos a 273 k y 1 atm se obtienen cuando reaccionan tatalmente 4,0 g de alumiinio ? ejercisio
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aplicar la ley de los gases ideales
P x V = n x R x T
T = 273 K
R = 0.0821 ( L atm / mol K)
Mm Al = 27 g/mol
V = ?
P = 1 atm
2Al + 6HCl → 2AlCl3 + 3H2
1. calcular moles n = masa/Mm
a. moles de Al
n Al = 4.0 g / 27 g/mol = 0.148 mol Al
n AlH2 = 0.148 mol Al x 3 mol H2
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2 mol Al
n H2 = 0.222 moles
2. calcular volumen
V = n x R x T
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P
V = 0.222 moles x 0.08/21 (L atm / mol K) x 273 K
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1 atm
V = 4.97 litros
P x V = n x R x T
T = 273 K
R = 0.0821 ( L atm / mol K)
Mm Al = 27 g/mol
V = ?
P = 1 atm
2Al + 6HCl → 2AlCl3 + 3H2
1. calcular moles n = masa/Mm
a. moles de Al
n Al = 4.0 g / 27 g/mol = 0.148 mol Al
n AlH2 = 0.148 mol Al x 3 mol H2
``````````````
2 mol Al
n H2 = 0.222 moles
2. calcular volumen
V = n x R x T
``````````````
P
V = 0.222 moles x 0.08/21 (L atm / mol K) x 273 K
``````````````````````````````````````````````````````````````````
1 atm
V = 4.97 litros
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