Answer:
0.169 mole of Al
Explanation:
We'll begin by by calculating the number of mole in 45 gof AlBr₃. This can be obtained as follow:
Mass of AlBr₃ = 45 g
Molar mass of AlBr₃ = 27 + (3×80)
= 27 + 240
= 267 g/mol
Mole of AlBr₃ =?
Mole = mass /molar mass
Mole of AlBr₃ = 45 / 267
Mole of AlBr₃ = 0.169 mole
Finally, we shall determine the number of mole of Al needed to produce 45 g (i.e 0.169 mole) of AlBr₃. This can be obtained as illustrated below:
2Al + 3Br₂ —> 2AlBr₃
From the balanced equation above,
2 moles of Al reacted to produce 2 moles of AlBr₃.
Therefore, 0.169 mole of Al will also react to produce 0.169 mole of AlBr₃.
Thus, 0.169 mole of Al is needed for the reaction.
I believe that it is 187.56 g/mol
Answer:
The system will try to balance the change by shifting toward the exothermic reaction, and the rate of the forward reaction will increase.
Explanation:
To balance the external cooling system has to give out heat so exothermic reaction will occur .
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