When Maximum number of of moles and gram can form, when 503 gm ICl3 react with 1

Question

When Maximum number of of moles and gram can form, when 503 gm ICl3 react with 170.8g of H2O ICl3 + H2O----- Icl + HIO3 + Hcl . What mass of the excess reactant remains?

Explanation / Answer

656 g iodine trichloride 233.2641 g/mol = 2.812 moles ICl3 115.6 g water @ 18.01 g/mol = 6.385 moles H2O by the equation 2ICl3 + 3H2O ---------> ICl + HIO3 + 5HCl we only needed 3/2 as many moles of H2O to react with ICl3 they added too much water, your limiting reagent is ICl3 by the equation 2ICl3 + 3H2O ---------> ICl + HIO3 + 5HCl 2.812 moles ICl3 produces 1/2 as many moles of HIO3 = 1.406 moles HIO3 find moles, using molar mass 1.406 moles HIO3 @ 175.91 g/mol = 247.3 grams of HIO3 Calculate the maximum number of moles and mass of iodic acid (HIO3) that can form your answers are 1.406 moles HIO3 247.3 grams of HIO3 ======================================

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