5.40 g of aluminium (Ar = 27.0) burns in excess chlorine: 2Al + 3Cl2 -> 2AlCl3. What mass of aluminium chloride (Mr = 133.5) forms?
Answer and explanation
Correct answer: D. 26.7 g
- Step 1: Moles Al . Step 2. Ratio Al : AlCl3 = 2 : 2 = 1 : 1, so moles AlCl3 . Step 3. Mass g. Answer: 26.7 g. Why not the others: 40.1 g applies a 3:2 ratio the wrong way; 17.8 g applies 2:3 the wrong way; 53.4 g doubles the correct answer.
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