Asked by Grace
0.27f of aluminium reacts with 2.4g of bromine. What is the relative atomic mass of a compound with this formula?
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Assume "0.27f" = 0.27 g.
Moles Al = 0.27 / 27 = 0.01 mol
Moles Br = 2.4 / 80 ≈ 0.03 mol
Ratio Al : Br = 0.01 : 0.03 = 1 : 3, so formula is AlBr3.
Relative formula (molecular) mass Mr ≈ Ar(Al) + 3×Ar(Br) ≈ 27 + 3×80 = 267 (more precisely ≈ 266.7 g·mol⁻¹).
Moles Al = 0.27 / 27 = 0.01 mol
Moles Br = 2.4 / 80 ≈ 0.03 mol
Ratio Al : Br = 0.01 : 0.03 = 1 : 3, so formula is AlBr3.
Relative formula (molecular) mass Mr ≈ Ar(Al) + 3×Ar(Br) ≈ 27 + 3×80 = 267 (more precisely ≈ 266.7 g·mol⁻¹).
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