The amount of KCl that would be formed will be 478.8 grams
Stoichiometric calculations
From the equation of the reaction, the mole ratio of K to [tex]Cl_2[/tex] is 2:1.
Mole of 309 g potassium = 309/39 = 7.92 moles
Mole of 228 g Cl2 = 228/71 = 3.21 moles
Thus, potassium is in excess while Cl2 is limiting.
Mole ratio of Cl2 and KCl = 1:2
Equivalent mole of KCl = 3.21 x 2 = 6.42 moles
Mass of 6.42 moles KCl = 6.42 x 74.55 = 478.8 grams
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