For the reaction
2C2H5OH + 2Na2Cr2O7 + 16HCI → 3CH3COOH + 4CrCl3 + 4NaCl + 11H2O
the yield (called a theoretical yield) of CrCl3 we can get from the reaction of 14.0 mol of C2H5OH, 10.0 mol of Na2Cr2O7 and 50.0 mol of HCI would be
(a) 20.0 mol
(b) 28.0 mol
(c) 10.0 mol
(d) 12.5 mol
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