Question

Solve this Stoichiometry problem

Answer 1

Answer : The mass of $Br_2$ needed are, 131.8 grams

Solution : Given,

Mass of Al = 15 g

Molar mass of Al = 27 g/mole

Molar mass of $Br_2$ = 159.8 g/mole

First we have to calculate the moles of Al.

$\text{Moles of Al}=\frac{\text{Mass of Al}}{\text{Molar mass of Al}}=\frac{15g}{27g/mole}=0.55mole$

Now we have to calculate the moles of $Br_2$

The given balanced reaction will be,

$2Al+3Br_2\rightarrow 2AlBr_3$

From the balanced reaction, we conclude that

As, 2 moles of Al react with 3 moles of $Br_2$

So, 0.55 mole of Al react with $\frac{3}{2}\times 0.55=0.825mole$ of $Br_2$

Now we have to calculate the mass of $Br_2$

$\text{Mass of }Br_2=\text{Moles of }Br_2\times \text{Molar mass of }Br_2$

$\text{Mass of }Br_2=0.825mole\times 159.8=131.8g$

Therefore, the mass of $Br_2$ needed are, 131.8 grams

Answer 2

Answer : For, this question,

The balanced equation is; 2Al + 3$Br_{2}$ ----> 2Al$Br_{3}$

We can now, calculate how many grams of will be needed to completely convert 15 g of Al into Al$Br_{3}$.

Here, the reaction produces, 2 moles of Al$Br_{3}[/tex} by consuming 2 moles of Al and 3 moles of [tex]Br_{2}$

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 $Br_{2}$ will be needed.

Therefore, 133g of will be needed to completely convert 15 g of Al into Al$Br_{3}$.