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Astudent Performs the Following Reaction: They Reacted 28.0 Grams of Lodine (V) I_(2)O_(5) with 80.0 Grams of Carbon Monoxide, CO. How

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Astudent performs the following reaction: They reacted 28.0 grams of lodine (V) I_(2)O_(5) with 80.0 grams of carbon monoxide, CO. How many grams of lodine. I_(2) did they produce? 1_(2)O_(5)+5COarrow 5CO_(2)+I_(2) 135.31 g 14498 338288 21.3g

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Respuesta

To solve this problem, we need to use stoichiometry to determine the amount of iodine (I2) produced from the given reaction.Given information:- 28.0 grams of iodine pentoxide (I2O5) is reacted with 800 grams of carbon monoxide (CO).- The balanced chemical equation is: I2O5 + 5CO → 5CO2 + I2.Step 1: Calculate the molar mass of I2O5.Molar mass of I2O5 = (2 × 126.90 g/mol) + (5 × 16.00 g/mol) = 333.80 g/molStep 2: Calculate the number of moles of I2O5.Moles of I2O5 = 28.0 g / 333.80 g/mol = 0.0843 molStep 3: Use the stoichiometry of the reaction to calculate the moles of I2 produced.According to the balanced equation, 1 mole of I2O5 produces 1 mole of I2.Moles of I2 produced = 0.0843 molStep 4: Calculate the mass of I2 produced.Mass of I2 = Moles of I2 × Molar mass of I2Molar mass of I2 = 253.81 g/molMass of I2 = 0.0843 mol × 253.81 g/mol = 21.3 gTherefore, the student produced 21.3 grams of iodine (I2) in the reaction.