Question

How many grams of Br2 are needed to form 67.1 g of AlBr3 ?

2Al(s)+3Br2(l)⟶2AlBr3(s)

Answer 1

Approximately
`60.3\; \rm g`.

Step-by-step explanation:

Look up the relative atomic mass of
`\rm Al` and
`\rm Br` on a modern periodic table:

• 
  `\rm Al`:
  `26.982`.
• 
  `\rm Br`:
  `79.904`.

Calculate the formula mass of
`\rm AlBr_3` and
`\rm Br_2`:

`\begin{aligned}& M(\mathrm{AlBr_3}) = 26.982 + 3 * 79.904 \approx 266.694\; \rm g \cdot mol^(-1) \\ & M(\mathrm{Br_2}) = 2* 79.904 \approx 159.808\; \rm g \cdot mol^(-1)\end{aligned}`.

Calculate the number of moles of formula units in
`67.1\; \rm g` of
`\rm AlBr_3`:

`\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{m(\mathrm{AlBr_3})}{M(\mathrm{AlBr_3})} \\ &\approx (67.1\; \rm g)/(266.694\; \rm g \cdot mol^(-1)) \approx 0.2516\; \rm mol \end{aligned}`.

Refer to the balanced equation for this reaction. The ratio between the coefficients of
`\rm Br_2` and
`\rm AlBr_3` in that equation is three-to-two. That corresponds to the ratio:

`\begin{aligned}\frac{n(\text{$\mathrm{Br_2}$, consumed})}{n(\text{$\mathrm{AlBr_3}$, produced})} &= (3)/(2)\end{aligned}`.

It is already calculated that approximately
`0.2516\; \rm mol` of
`\rm AlBr_3` was produced through this reaction. Apply this ratio to approximate the (minimum) number of moles of
`\rm Br_2` that is consumed:

`\displaystyle (3)/(2) * 0.2516\; \rm mol \approx 0.3774\; \rm mol`.

Calculate the mass of that
`0.3774\; \rm mol` of
`\rm Br_2`:

`\begin{aligned}m(\mathrm{Br_2}) &= n(\mathrm{Br_2})\cdot M(\mathrm{Br_2}) \\ &\approx 0.3774\; \rm mol * 159.808\; \rm g \cdot mol^(-1) \approx 60.3\; \rm g\end{aligned}`.