4Al(s) + 3O2(g) --> 2Al2O3(s) This is the balanced.
From the equation:
4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3
3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3 (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3
4.5 moles of O2 reacted with 6moles of Al to produce 3moles of Al2O3
(3) is the answer. 6 mol of Al.
Answer:
0.145 moles de AlBr3.
Explanation:
¡Hola!
En este caso, al considerar la reacción química dada:
Al(s)+Br2(l)⟶AlBr3(s)
Es claro que primero debemos balancearla como se muestra a continuación:
2Al(s)+3Br2(l)⟶2AlBr3(s)
Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.
¡Saludos!
No. The only thing that changed was the looks of the gasoline, not the chemical components.
Psychical(you see the changes visible) and Chemical properties(they react and combine, you may see fizzing or bubbles).