Question

How many grams of Br2 are needed to form 67.1 g of AlBr3 ? 2Al(s)+3Br2(l)⟶2AlBr3(s)

Answer 1

Answer:

Approximately $60.3\; \rm g$.

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 $\rm Br_2$:

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

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

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

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:

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

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:

$\displaystyle \frac{3}{2} \times 0.2516\; \rm mol \approx 0.3774\; \rm mol$.

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

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