Question

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.

God bless

Answer 1

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.

Solution. We find the amount of iron phosphate: $n(\mathrm{FePO}_4) = \frac{m(\mathrm{FePO}_4)}{M(\mathrm{FePO}_4)}$, where $m(\mathrm{FePO}_4)$ - mass of iron phosphate, g; $M(\mathrm{FePO}_4)$ - molar weight of iron phosphate, g/mol. Then $n(\mathrm{FePO}_4) = \frac{54}{151} = 0.36$ mol. By the reaction equation at 100% reaction yield $n(\mathrm{FePO}_4) = n(\mathrm{Fe(NO}_3)_3)$. In this case, we have a yield of 90%, then $n(\mathrm{Fe(NO}_3)_3) = \frac{n(\mathrm{FePO}_4)}{0.9} = \frac{0.36}{0.9} = 0.4$ mol. The mass of iron nitrate is calculated as: $m(\mathrm{Fe(NO}_3)_3) = n(\mathrm{Fe(NO}_3)_3) \times M(\mathrm{Fe(NO}_3)_3) = 0.4 \times 242 = 96.8$ g.

Answer: 96.8 g.

Answer provided by AssignmentExpert.com