Aluminum will react with bromine to form aluminum bromide (Al2Br6). What mass of bromine (g) is needed to form 1.1 mol of Al2Br6? Plz help!?
回答 (1)
2017-10-18 4:55 pm
2Al + 3Br₂ → Al₂Br₆
Mole ratio Br₂ : Al₂Br₆ = 3 : 1
No. of moles of Al₂Br₆ formed = 1.1 mol
No. of moles of Br₂ needed = (1.1 mol) × 3 = 3.3 mol
Molar mass of Br₂ = 79.9 × 2 g/mol = 159.8 g/mol
Mass of Br₂ needed = (3.3 mol) × (159.8 g/mol) = 527 g
Mole ratio Br₂ : Al₂Br₆ = 3 : 1
No. of moles of Al₂Br₆ formed = 1.1 mol
No. of moles of Br₂ needed = (1.1 mol) × 3 = 3.3 mol
Molar mass of Br₂ = 79.9 × 2 g/mol = 159.8 g/mol
Mass of Br₂ needed = (3.3 mol) × (159.8 g/mol) = 527 g
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