Consider the following reaction: steel (s) + chlorine (g) à stole (III) chloride a. Compose the balanced chemical equation. B. 25.0 g of iron reacts through excess chlorine gas. A. Calculation the moles of steel reactant. B. Calculate the mole of stole (III) chloride. C. Calculate the massive of stole (III) chloride.

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

2Fe + 3Cl2 →2FeCl3

25 grams of steel is 25/55.84 moles of iron = 0.448 moles. according to the above balanced equation, 2 moles of iron produce 2 mole of Ferric Chloride, a 1 : 1 ratio. Therefore, the 0.448 mole of iron will certainly react v chlorine to develop 0.448 moles of iron+++ chloride. The formula load of FeCl3 is 55.84 + 3 x 35.45 = 162.19. Due to the fact that 0.448 mole of product to be formed, the massive of the product is 0.448 moles x 162.19 grams/mole = 72.66 grams.

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