At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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 :

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].
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.