Question

Solve this Stoichiometry problem

2 2 unread replies. 2 2 replies.
Solve the following problem showing all of your work and including all units. You can solve the problem in several steps or as a tandem conversion problem.

How many grams of Br2 are needed to completely convert 15.0 g Al to AlBr3 ?

Al (s) + Br2 (l) --> AlBr3 (hint: the equation is not balanced)

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}}=(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
`(3)/(2)* 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* \text{Molar mass of }Br_2`

`\text{Mass of }Br_2=0.825mole* 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)`.