Calculate the maximum numbers of moles and grams of iodic acid (HIO3 ) that can form when 376 g of iodine trichloride reacts with 225.1 g of water: ICl3 + H2O → ICl + HIO3 + HCl [unbalanced] ____ mol HIO3 _____ g HIO3 What mass of the excess reactant remains? ____g
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Step 1
806 moles - Moles of H2O = 12.5 moles From the balanced equation, the mole ratio between ICl3 and HIO3 is 1:1. Therefore, the limiting reactant is ICl3. Moles of HIO3 = 2.806 moles ICl3 * (1 mole HIO3 / 2 moles ICl3) = 1.403 moles HIO3 Show more…
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