1-¿Cuantas moles de oxido de aliminio se produciria a partir de 0.5 mol de oxigeno?
4AI+302 ->2AI203
2- ¿Cuantas moles de hidroxido de aluminio se requiere para producir 22 moles de carga? 2AI(OH)+3H2SO4->AI2SO4)3+6H2O
Respuestas
1- ¿Cuantas moles de oxido de aliminio se produciria a partir de 0.5 mol de oxigeno?
4AI + 3O2 → 2AI2O3
n Al2O3 = 0.5 mol O2 x 2 mol Al2O3
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3 mol O2
n Al2O3 = 0.33 moles
2- ¿Cuantas moles de hidroxido de aluminio se requiere para producir 22 moles de agua?
2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
n AI(OH)3 = 22 mol H2O x 2 mol AI(OH)3
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6 mol H2O
n Al(OH)3 = 7.33 moles
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2Al(OH)3 ↔ Al2O3 + 3H2O
n AI(OH)3 = 22 mol H2O x 2 mol AI(OH)3
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3 mol H2O
n AI(OH)3 = 14.67 moles