Respuesta :
The reactant has excess aluminum so bromine would be the limiting reagent. Looking at the coefficient of the equation, every 3 mol Br2 will produce 2 mol of AlBr3.
So, the amount of AlBr3 produced by 6 mol of Br2 would be:
Br2/AlBr3= 3/2
6mol/AlBr3= 3/2
AlBr3= 6mol * 2/3= 4 mol
So, the amount of AlBr3 produced by 6 mol of Br2 would be:
Br2/AlBr3= 3/2
6mol/AlBr3= 3/2
AlBr3= 6mol * 2/3= 4 mol
Answer : The number of moles of [tex]AlBr_3[/tex] formed will be, 4 moles
Explanation : Given,
Moles of [tex]Br_2[/tex] = 6.0 mole
The given balanced chemical reaction is,
[tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]
From the balanced chemical reaction, we conclude that
As, 3 moles of [tex]Br_2[/tex] react to give 2 moles of [tex]AlBr_3[/tex]
So, 6 moles of [tex]Br_2[/tex] react to give [tex]\frac{2}{3}\times 6=4[/tex] moles of [tex]AlBr_3[/tex]
Therefore, the number of moles of [tex]AlBr_3[/tex] formed will be, 4 moles
