If 0.073g of aluminum are used how many molecules of chlorine will be needed?

Answer 1

First balance the synthesis reaction to get mole ratio: 2 Al + 3 Cl₂ --> 2 AlCl₃.

Stoichiometry = dimensional analysis: (0.073g Al) * (1 mole Al /26.98g Al) * (3 mole Cl₂ /2 mole Al) = 0.0040₆ mole Cl₂ gas. Avogadro's Number: (6.022 x 10²³ molecules/mole) * (0.0040₆ mole) = 2.4 x 10²¹ molecules of chlorine.