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].
