Fe(NO3)3 (aq) + Na3PO4 (aq) ——> FePO4 + 3NaNO3 (aq)
assuming a percentage yield of 90 percent, what mass of ferric nitrate must be reacted with an excess of sodium phosphate to produce 54 g of ferric phosphate? The MWs are Fe(NO3)3 = 242 g, and FePO4 is 151 g.
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