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JH

# Like many metals, aluminum reacts with a halogen to give a metal halide (see Figure 3.1 ).$$2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Br}_{2}(\ell) \longrightarrow \mathrm{Al}_{2} \mathrm{Br}_{6}(\mathrm{s})$$ What mass of $\mathrm{Br}_{2}$, in grams, is required for complete reaction with $2.56 \mathrm{g}$ of Al? What mass of white, solid $\mathrm{Al}_{2} \mathrm{Br}_{6}$ is expected?

## $22.7 \mathrm{~g} \mathrm{Br}_{2} ; 25.3 \mathrm{~g} \mathrm{Al}_{2} \mathrm{Br}_{6}$

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### Video Transcript

and this problem were asked How much bro mean Ingram's is needed to react with 2.56 grams of aluminum. The first thing we're gonna do is to convert the 2.56 grams of aluminum into the corresponding number of moles by dividing it by its atomic mass. Now that we know we need sarah 0.9 for nine moles of aluminum, we also concede e that from the balance reaction, three moles of bro mean are required per two moles of aluminum. From that, we can calculate that we need 22.7 grams of aluminum to complete this reaction and the second part of this problem were asked how much product is expected knowing that we get one mole of aluminum bromide for two moles of aluminum, we can calculate this by dividing 0.949 by two and multiplying that by the molar mass of aluminum. Bro, my to get 25.3 grams of product produced

JH
Brown University

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