Aluminum reacts with bromine to form aluminum bromide in the reaction: 2Al + 3Br2 → 2AlBr3 If 15.79 g of aluminum reacts with 55.61 g of bromine, how many grams of aluminum bromide will form? 1. 78.01 g 2. 156.02 g 3. 123.75 g 4. 30.94 g 5. 61.87 g?
The Correct Answer and Explanation is:
To determine how many grams of aluminum bromide (AlBr₃) will form, we need to use stoichiometry, which involves converting grams to moles, using the molar ratio from the balanced equation, and then converting the result back to grams.
Step 1: Write the balanced equation
The balanced equation is:
2Al+3Br2→2AlBr32Al + 3Br_2 \rightarrow 2AlBr_32Al+3Br2→2AlBr3
This shows that 2 moles of aluminum (Al) react with 3 moles of bromine (Br₂) to produce 2 moles of aluminum bromide (AlBr₃).
Step 2: Convert grams to moles for both reactants
- Molar mass of aluminum (Al) = 26.98 g/mol
- Molar mass of bromine (Br₂) = 159.8 g/mol (since the molar mass of Br is 79.9 g/mol)
Moles of aluminum:
Moles of Al=Mass of AlMolar mass of Al=15.79 g26.98 g/mol=0.585 mol Al\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{15.79 \, \text{g}}{26.98 \, \text{g/mol}} = 0.585 \, \text{mol Al}Moles of Al=Molar mass of AlMass of Al=26.98g/mol15.79g=0.585mol Al
Moles of bromine:
Moles of Br2=Mass of Br2Molar mass of Br2=55.61 g159.8 g/mol=0.348 mol Br2\text{Moles of Br}_2 = \frac{\text{Mass of Br}_2}{\text{Molar mass of Br}_2} = \frac{55.61 \, \text{g}}{159.8 \, \text{g/mol}} = 0.348 \, \text{mol Br}_2Moles of Br2=Molar mass of Br2Mass of Br2=159.8g/mol55.61g=0.348mol Br2
Step 3: Identify the limiting reactant
The balanced equation tells us that the molar ratio between Al and Br₂ is 2:3. Let’s compare the ratio of the available moles of each reactant:0.585 mol Al2=0.2925\frac{0.585 \, \text{mol Al}}{2} = 0.292520.585mol Al=0.29250.348 mol Br23=0.116\frac{0.348 \, \text{mol Br}_2}{3} = 0.11630.348mol Br2=0.116
Since 0.116 mol of Br₂ is less than 0.2925 mol of Al, bromine (Br₂) is the limiting reactant.
Step 4: Use the limiting reactant to calculate the moles of AlBr₃ produced
From the balanced equation, we know that 3 moles of Br₂ produce 2 moles of AlBr₃. Therefore, we can calculate the moles of AlBr₃ produced from 0.348 mol of Br₂:Moles of AlBr3=2 mol AlBr33 mol Br2×0.348 mol Br2=0.232 mol AlBr3\text{Moles of AlBr}_3 = \frac{2 \, \text{mol AlBr}_3}{3 \, \text{mol Br}_2} \times 0.348 \, \text{mol Br}_2 = 0.232 \, \text{mol AlBr}_3Moles of AlBr3=3mol Br22mol AlBr3×0.348mol Br2=0.232mol AlBr3
Step 5: Convert moles of AlBr₃ to grams
The molar mass of AlBr₃ is:Molar mass of AlBr3=26.98 g/mol+3×79.9 g/mol=266.68 g/mol\text{Molar mass of AlBr}_3 = 26.98 \, \text{g/mol} + 3 \times 79.9 \, \text{g/mol} = 266.68 \, \text{g/mol}Molar mass of AlBr3=26.98g/mol+3×79.9g/mol=266.68g/mol
Now, we can calculate the mass of AlBr₃:Mass of AlBr3=Moles of AlBr3×Molar mass of AlBr3=0.232 mol×266.68 g/mol=61.87 g\text{Mass of AlBr}_3 = \text{Moles of AlBr}_3 \times \text{Molar mass of AlBr}_3 = 0.232 \, \text{mol} \times 266.68 \, \text{g/mol} = 61.87 \, \text{g}Mass of AlBr3=Moles of AlBr3×Molar mass of AlBr3=0.232mol×266.68g/mol=61.87g
Final Answer:
The mass of aluminum bromide (AlBr₃) that will form is 61.87 g.
Thus, the correct answer is 5. 61.87 g.
