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How many grams of [tex]\[Br_2\][/tex] are needed to form 72.1 g of [tex]\[AlBr_3\][/tex]?

[tex]\[
2Al(s) + 3Br_2(l) \longrightarrow 2AlBr_3(s)
\][/tex]

Step 1: Show the strategy for solving this problem.
[tex]\[ \text{grams } AlBr_3 \longrightarrow \][/tex]
[tex]\[ \text{moles } AlBr_3 \longrightarrow \][/tex]
[tex]\[ \text{moles } Br_2 \longrightarrow \][/tex]
[tex]\[ \text{grams } Br_2 \][/tex]

Answer Bank:
- grams [tex]\[Br_2\][/tex]
- moles [tex]\[AlBr_3\][/tex]
- moles [tex]\[Br_2\][/tex]
- grams [tex]\[AlBr_3\][/tex]
- moles Al
- grams Al

Sagot :

GinnyAnsweridn
To solve the problem of finding how many grams of [tex]\( Br_2 \)[/tex] are needed to form 72.1 grams of [tex]\( AlBr_3 \)[/tex], we can follow this step-by-step strategy:

1. Convert grams of [tex]\( AlBr_3 \)[/tex] to moles of [tex]\( AlBr_3 \)[/tex]:
[tex]\[ \text{grams } AlBr_3 \longrightarrow \text{moles } AlBr_3 \][/tex]

2. Use stoichiometric coefficients from the balanced chemical equation to convert moles of [tex]\( AlBr_3 \)[/tex] to moles of [tex]\( Br_2 \)[/tex]:
[tex]\[ \text{moles } AlBr_3 \longrightarrow \text{moles } Br_2 \][/tex]

3. Convert moles of [tex]\( Br_2 \)[/tex] to grams of [tex]\( Br_2 \)[/tex]:
[tex]\[ \text{moles } Br_2 \longrightarrow \text{grams } Br_2 \][/tex]

Now, let's go through these steps:

### Step 1: Convert grams of [tex]\( AlBr_3 \)[/tex] to moles of [tex]\( AlBr_3 \)[/tex].
Given:
- [tex]\( \text{molar mass of AlBr}_3 = 26.98 + 3 \times 79.904 = 266.692 \, \text{g/mol} \)[/tex]
- [tex]\( \text{mass of AlBr}_3 = 72.1 \, \text{g} \)[/tex]

Calculate moles of [tex]\( AlBr_3 \)[/tex]:

[tex]\[ \text{moles } AlBr_3 = \frac{\text{grams } AlBr_3}{\text{molar mass of } AlBr_3} = \frac{72.1 \, \text{g}}{266.692 \, \text{g/mol}} \approx 0.270 \, \text{mol} \][/tex]

### Step 2: Convert moles of [tex]\( AlBr_3 \)[/tex] to moles of [tex]\( Br_2 \)[/tex].
From the balanced chemical equation, we know:

[tex]\[ 2 \, AlBr_3 \longrightarrow 3 \, Br_2 \][/tex]

Thus, 1 mole of [tex]\( AlBr_3 \)[/tex] produces 1.5 moles of [tex]\( Br_2 \)[/tex].

[tex]\[ \text{moles } Br_2 = \text{moles } AlBr_3 \times 1.5 = 0.270 \, \text{mol} \times 1.5 = 0.4055 \, \text{mol} \][/tex]

### Step 3: Convert moles of [tex]\( Br_2 \)[/tex] to grams of [tex]\( Br_2 \)[/tex].
Given:
- [tex]\( \text{molar mass of Br}_2 = 2 \times 79.904 = 159.808 \, \text{g/mol} \)[/tex]

Calculate grams of [tex]\( Br_2 \)[/tex]:

[tex]\[ \text{grams } Br_2 = \text{moles } Br_2 \times \text{molar mass of } Br_2 = 0.4055 \, \text{mol} \times 159.808 \, \text{g/mol} = 64.806 \, \text{g} \][/tex]

So, approximately 64.806 grams of [tex]\( Br_2 \)[/tex] are needed to form 72.1 grams of [tex]\( AlBr_3 \)[/tex].
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