Asked by Grace
0.27f of aluminium reacts with 2,4g of bromine. Write a possible formula for the compound of aluminium with bromine which is produced
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Moles Al = 0.27 g / 27.0 g·mol⁻¹ = 0.0100 mol
Moles Br atoms = 2.4 g / 79.9 g·mol⁻¹ ≈ 0.0300 mol
Ratio Br : Al = 0.0300 : 0.0100 = 3 : 1
Empirical formula = AlBr3 (note: molecular form may be dimeric Al2Br6, but the formula unit is AlBr3).
Moles Br atoms = 2.4 g / 79.9 g·mol⁻¹ ≈ 0.0300 mol
Ratio Br : Al = 0.0300 : 0.0100 = 3 : 1
Empirical formula = AlBr3 (note: molecular form may be dimeric Al2Br6, but the formula unit is AlBr3).
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