Answer:
0.109 moles Al
Explanation:
To find the necessary amount of aluminum (Al), you need to (1) convert grams AlBr₃ to moles AlBr₃ (via molar mass from periodic table values) and then (2) convert moles AlBr₃ to moles Al (via coefficients from reaction). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs because the given value (29.0) has 3 sig figs.
Molar Mass (AlBr₃): 26.982 g/mol + 3(79.904 g/mol)
Molar Mass (AlBr₃): 266.694 g/mol
2 Al(s) + 3 Br₂(l) --> 2 AlBr₃(s)
29.0 g AlBr₃ 1 mole 2 moles Al
------------------ x --------------------- x ----------------------- = 0.109 moles Al
266.694 g 2 moles AlBr₃
