Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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 :

GinnyAnswer
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].
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.