Question

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

Answer 1

To determine the mass of the product, we need to determine the limiting reactant in the reaction of aluminum and bromine. By comparing the moles of the reactants, we can determine which one is the limiting reactant and then calculate the mass of the product using the molar mass.

Step-by-step explanation:

To determine the mass of product that can be made, we need to determine the limiting reactant in the reaction. The balanced equation for the reaction is 2 Al + 3 Br2 → 2 AlBr3. From the given masses, we can calculate the moles of aluminum and bromine. The limiting reactant is the one that produces the fewer moles of product, and the mass of the product can be calculated using the molar ratio of the product to the limiting reactant.

Let's calculate the moles of aluminum and bromine:

• Moles of aluminum = mass of aluminum / molar mass of aluminum
• Moles of bromine = mass of bromine / molar mass of bromine

Next, we compare the moles of the reactants to determine the limiting reactant:

• Aluminum: moles of aluminum × 2 moles AlBr3 / 2 moles Al = moles of AlBr3 formed
• Bromine: moles of bromine × 2 moles AlBr3 / 3 moles Br2 = moles of AlBr3 formed

The reactant that produces the smaller number of moles of AlBr3 is the limiting reactant. Finally, we can calculate the mass of the product using the moles of AlBr3 formed and the molar mass of AlBr3. Mass of AlBr3 = moles of AlBr3 × molar mass of AlBr3.

Answer 2

Approximately
`24\; \rm g` (at most.)

Step-by-step explanation:

Aluminum
`\rm Al` reacts with bromine
`\rm Br_2` at a
`2:3` ratio:

`\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)`.

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

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

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

• 
  `M(\mathrm{Al}) = 26.986\; \rm g \cdot mol^(-1)`.
• 
  `M(\mathrm{Br_2}) = 2* 79.904 = 159.808\; \rm g \cdot mol^(-1)`.
• 
  `M(\mathrm{AlBr_3}) = 26.986 + 3 * 79.904 = 266.698\; \rm g \cdot mol^(-1)`.

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

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

Assume that
`\rm Al\, (s)` is the limiting reactant. From the coefficients:

`\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1`.

Based on the assumption that
`\rm Al\, (s)` is the limiting reactant:

`\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1* 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}`.

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

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

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

Based on the assumption that
`\rm Br_2\, (g)` is the limiting reactant:

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

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

Apply the formula mass of
`\rm AlBr_3` to find the mass of that (approximately)
`0.0918\; \rm mol` of
`\rm AlBr_3` formula units:

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