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 follow these steps:
Step 1: Write the balanced chemical equation.
The reaction between aluminum (Al) and bromine (Br2) to form aluminum bromide is:
2Al + 3Br2 → 2AlBr3
Step 2: Find the molar masses of aluminum, bromine, and aluminum bromide.
- Molar mass of Al = 26.98 g/mol
- Molar mass of Br = 79.90 g/mol
- Molar mass of AlBr3 = (26.98 g/mol) + 3 × (79.90 g/mol) = 266.68 g/mol
Step 3: Convert the masses of aluminum and bromine to moles.
We are given:
- Mass of Al = 15.79 g
- Mass of Br2 = 55.61 g
For aluminum:moles of Al=15.79 g26.98 g/mol=0.585 mol\text{moles of Al} = \frac{15.79 \, \text{g}}{26.98 \, \text{g/mol}} = 0.585 \, \text{mol}moles of Al=26.98g/mol15.79g=0.585mol
For bromine:moles of Br2=55.61 g159.80 g/mol=0.348 mol\text{moles of Br2} = \frac{55.61 \, \text{g}}{159.80 \, \text{g/mol}} = 0.348 \, \text{mol}moles of Br2=159.80g/mol55.61g=0.348mol
Step 4: Determine the limiting reactant.
From the balanced equation, we know:
- 2 moles of Al react with 3 moles of Br2.
Thus, the mole ratio of Al to Br2 is 2:3. To find the limiting reactant, we need to check which reactant will run out first.
For 0.585 moles of Al, we need:moles of Br2=32×0.585=0.878 mol\text{moles of Br2} = \frac{3}{2} \times 0.585 = 0.878 \, \text{mol}moles of Br2=23×0.585=0.878mol
But we only have 0.348 moles of Br2, which is less than 0.878 mol, so bromine (Br2) is the limiting reactant.
Step 5: Calculate the amount of AlBr3 produced.
From the balanced equation:
- 3 moles of Br2 produce 2 moles of AlBr3.
So, 0.348 moles of Br2 will produce:moles of AlBr3=23×0.348=0.232 mol\text{moles of AlBr3} = \frac{2}{3} \times 0.348 = 0.232 \, \text{mol}moles of AlBr3=32×0.348=0.232mol
Now, we can calculate the mass of AlBr3 produced:mass of AlBr3=0.232 mol×266.68 g/mol=61.87 g\text{mass of AlBr3} = 0.232 \, \text{mol} \times 266.68 \, \text{g/mol} = 61.87 \, \text{g}mass of AlBr3=0.232mol×266.68g/mol=61.87g
Final Answer:
61.87 g of aluminum bromide will form. Therefore, the correct option is:
5. 61.87 g.
