Answer:
The mass of AlBr3 is 24.5 grams
Explanation:
Step 1: Data given
Mass of aluminium = 5.0 grams
Mass of bromine = 22.0 grams
Molar mass of aluminium = 26.98 g/mol
Molar mass of br2= 159.8 g/mol
Step 2: The balancced equation
2 Al + 3 Br2 → 2 AlBr3
Step 3: Calculate moles Al
Moles Al = mass Al / molar mass Al
Moles Al = 5.0 grams / 26.98 g/mol
Moles Al = 0.185 moles
Step 4: Calculate moles Br
Moles Br = 22.0 grams / 159.8 g/mol
Moles Br = 0.138 moles
Step 5: Calculate limiting reactant
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
Br2 is the limiting reactant. It will completely be consumed (0.0313 moles)
Al is in excess. There will react 2/3*0.138 = 0.092 moles
There will remain 0.185- 0.092 = 0.093 moles Al
Step 6: Calculate moles AlBr3
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
For 0.0313 moles Br2 we'll have 2/3*0.138 = 0.092 moles AlBr3
Step 7: Calculate mass AlBr3
Mass AlBr3 = moles * molar mass
Mass AlBr3 = 0.092 moles * 266.69 g/mol
Mass AlBr3 = 24.5 grams
The mass of AlBr3 is 24.5 grams
= 5.00 mol O₂