What is the maximum mass of product that can be formed when 10.5 g of aluminum and 23.5 g fluorine react?
From equation 2x 27 54g Al react with 6 x 19 114g fluorine [use exact atomic mass given by your teacher -I've rounded to 19] forming 168g AlF3 so 1g Al reacts with 114/54g fluorine forming 168/54g AlF3 so 10.5g Al reacts with 10.5 x 114/54g fluorine 22.17g so we have more than enough fluorine, so Al is the limiting reagent here so 10.5g Al reacts with 10.5 x 114/54g fluorine forming 10.5 x 168/54g AlF3 32.67g answer AlF3 max formed 32.67g
