What mass of product, in grams, can be made by reacting 5.0g of aluminum and 22g of bromine?

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25-07-2020
Chemistry

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What mass of product, in grams, can be made by reacting 5.0g of aluminum and 22g of bromine?

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jacob193 jacob193 · Answer 1 · 2020-07-25T21:20:27+02:00

Answer:

Approximately [tex]24\; \rm g[/tex] (at most.)

Explanation:

Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:

[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].

Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:

[tex]\rm Al[/tex]: [tex]26.982[/tex].
[tex]\rm Br[/tex]: [tex]79.904[/tex].

Calculate the formula mass of the reactants and of the product:

[tex]M(\mathrm{Al}) = 26.986\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{Br_2}) = 2\times 79.904 = 159.808\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{AlBr_3}) = 26.986 + 3 \times 79.904 = 266.698\; \rm g \cdot mol^{-1}[/tex].

Calculate the quantity (in number of moles of formula units) of each reactant:

[tex]\displaystyle n(\mathrm{Al}) = \frac{m(\mathrm{Al})}{M(\mathrm{Al})} = \frac{5.0\; \rm g}{26.986\; \rm g \cdot mol^{-1}} \approx 0.18528\; \rm mol[/tex].
[tex]\displaystyle n(\mathrm{Br_2}) = \frac{m(\mathrm{Br_2})}{M(\mathrm{Br_2})} = \frac{22\; \rm g}{159.808\; \rm g \cdot mol^{-1}} \approx 0.13767\; \rm mol[/tex].

Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].

Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].

In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.

On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].

Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].

Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].

Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].