Consider the following reaction: iron (s) + chlorine (g) à iron (III) chloride a. Write the balanced chemical equation. b. 25.0 g of iron reacts with excess chlorine gas. a. Calculate the moles of iron reactant. b. Calculate the moles of iron (III) chloride. c. Calculate the mass of iron (III) chloride.

You are watching: Solid iron reacts with chlorine gas to form solid iron(iii) chloride. Iron reacts with chlorine to form iron+++ Cl3 according to the following balanced equation:

2Fe + 3Cl2 →2FeCl3

25 grams of iron is 25/55.84 moles of iron = 0.448 moles. According to the above balanced equation, 2 moles of iron produce 2 moles of Ferric Chloride, a 1 : 1 ratio. Therefore, the 0.448 moles of iron will react with chlorine to produce 0.448 moles of iron+++ chloride. The formula weight of FeCl3 is 55.84 + 3 x 35.45 = 162.19. Since 0.448 moles of product were formed, the mass of the product is 0.448 moles x 162.19 grams/mole = 72.66 grams.

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