How many moles of Al are necessary to form 46.2 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Answer 1

Hello Bryanna,

To solve this problem, we must first find the moles of AlBr₃. To convert from 46.2g AlBr₃ to moles of AlBr₃, we simply divide by the molar mass of AlBr₃, which is 266.69 g/mol.

46.2g AlBr₃ x 1 molAlBr₃/266.69gAlBr₃= 0.173 moles AlBr₃

To go from moles of AlBr₃ to moles of Al, we simply multiply by the ratio of coefficients in the reaction. For every 2 mol of Al, we produce 2 mol AlBr_3. Therefore, our ratio is 2:2, or 1:1.

0.173 mol AlBr₃ x 2 mol Al/2 mol AlBr₃= 0.173 mol Al

Hope this helps!