Answer : For, this question,
The balanced equation is; 2Al + 3[tex]Br_{2}[/tex] ----> 2Al[tex]Br_{3}[/tex]
We can now, calculate how many grams of will be needed to completely convert 15 g of Al into Al[tex]Br_{3}[/tex].
Here, the reaction produces, 2 moles of Al[tex]Br_{3}[/tex} by consuming 2 moles of Al and 3 moles of [tex]Br_{2}[/tex]
So, For 15.0 g of Al; we can calculate;
(15.0 g Al) X (1 mole Al / 26.98 g Al) X (3 moles / 2 moles Al) X (159.81 g / 1 mole )
= 133 g of [tex]Br_{2}[/tex] will be needed.
Therefore, 133g of will be needed to completely convert 15 g of Al into Al[tex]Br_{3}[/tex].
Answer : The mass of [tex]Br_2[/tex] needed are, 131.8 grams
Solution : Given,
Mass of Al = 15 g
Molar mass of Al = 27 g/mole
Molar mass of [tex]Br_2[/tex] = 159.8 g/mole
First we have to calculate the moles of Al.
[tex]\text{Moles of Al}=\frac{\text{Mass of Al}}{\text{Molar mass of Al}}=\frac{15g}{27g/mole}=0.55mole[/tex]
Now we have to calculate the moles of [tex]Br_2[/tex]
The given balanced reaction will be,
[tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]
From the balanced reaction, we conclude that
As, 2 moles of Al react with 3 moles of [tex]Br_2[/tex]
So, 0.55 mole of Al react with [tex]\frac{3}{2}\times 0.55=0.825mole[/tex] of [tex]Br_2[/tex]
Now we have to calculate the mass of [tex]Br_2[/tex]
[tex]\text{Mass of }Br_2=\text{Moles of }Br_2\times \text{Molar mass of }Br_2[/tex]
[tex]\text{Mass of }Br_2=0.825mole\times 159.8=131.8g[/tex]
Therefore, the mass of [tex]Br_2[/tex] needed are, 131.8 grams
