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

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