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What is the mass of excess reactant leftover if 20.0 grams of chromium reacts with 10.0 grams of copper II sulfate in the BALANCED reaction below?
- Roger the MoleLv 72 months agoFavorite Answer
(20.0 g Cr) / (51.9961 g Cr/mol) = 0.384644 mol Cr
(10.0 g CuSO4) / (159.6086 g CuSO4/mol) = 0.0626533 mol CuSO4
0.0626533 mole of CuSO4 would react completely with 0.0626533 x (2/3) = 0.0417689 mole of Cr, but there is more Cr present than that, so Cr is in excess.
((0.384644 mol Cr initially) - (0.0417689 mol Cr reacted)) x (51.9961 g Cr/mol) =
17.8 g Cr left over