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3.3 mol Fe(OH)3 and 6.3 mole H2SO4react according to the equation 2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + 6H2O.
If the limiting reactant is Fe(OH)3, determine the amount of excess reactant that remains. Answer in units of mol.
2 mol of Fe(OH)3 produces 1 mol of Fe2(SO4)3 then 2.1 mol of Fe(OH)3 will produce (1)/(2) x 2.01 = 1.05 mol of Fe2(SO4)3 amount of H2O = (6)/(2) * 2.01 = 6.03 mol HO2
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