Question
When 80.5 mL of 0.642 M Ba(NO3)2 are mixed with 44.5 mL of 0.743 M KOH, a precipitate of Ba(OH)2 forms. Determine the limiting reactant,
When 80.5 mL of 0.642 M Ba(NO3)2 are mixed with 44.5 mL of 0.743 M KOH, a precipitate of Ba(OH)2 forms. Determine the limiting reactant, and the grams of precipitate formed. The balanced chemical equation for the reaction is: Ba(NO3)2(aq) + 2 KOH(aq) -> 2 KNO3(aq) + Ba(OH)2(s)
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Physical Chemistry
Authors: Peter Atkins
7th Edition
978-0716735397, 716735393, 978-0716743880
