The reaction of NbCl 4 (THF) 2 with pyridine in the presence of a reducing agent gives
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The reaction of NbCl4(THF)2 with pyridine in the presence of a reducing agent gives NbClx(py)y which contains 50.02% C, 4.20% H and 11.67% N. Determine the values of x and y.
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To determine the values of x and y in the compound NbClxpyy we need to calculate the molar ratios of Nb Cl py pyridine C H and N based on the given pe...View the full answer
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