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.
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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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