2Al + 3Br2 → 2AlBr3If 25.0 g of Al and 100g of Br2 react, and 61.8 g AlBr3 is recovered, what is the percent yield for the reaction?_%
use molar masses to find mmoles: 25.0 g of Al 26.98 g/mol 0.9266 moles of Al and 100g of Br2 159.8 g/mol 0.626 moles of Br2 because they failed to add the 50% more moles of Br2 than moles of Al, Br2 is the limiting reagent by the reaction: 2Al + 3Br2 → 2AlBr3 0.626 moles of Br2 produces 2/3rds as many moles of AlBr3 0.417 mol AlBr3 using molar mass, find grams that theoretically could be produced: 0.417 mol AlBr3 266.69 g/mol 111.26 g of AlBr3 find % yield: 61.8 g AlBr3 / 111.26 g of AlBr3 times 100 55.54% yield your answer, rounded to 3 sig figs is 55.5 % Yield
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