Atomic weights: F=18.998, Al=26.982, 2AlF₃=2(26.982+3×18.998)=167.953, 2Al=53.963, 3F₂=113.990 (approx values).
53.963/167.953=0.3213 or 32.13% of AlF₃ and is the proportion by weight of the amount of Al produced compared to the amount of AlF₃. So, starting with 2.1 moles we get 2.1×0.3213=0.6747 moles of Al (about 2.1/3=0.7).