Explanation:
We are given: mass of FePO4 = 15g
: mass of Na2SO4 = 5g
We first find the mass of Fe2(SO4)3 from the mass of FePO4:
m is the mass and M is the molar mass
[tex]\begin{gathered} m\text{ = }\frac{m(FePO4)}{M(FePO4)}\times\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{2mol\text{ FePO4}}\times\text{ M\lparen Fe2\lparen SO4\rparen3\rparen} \\ \text{ = }\frac{15}{150.82}\times\frac{1}{2}\times399.88 \\ \text{ = 19.89g} \end{gathered}[/tex]
We then find the mass of Fe2(SO4)3 from Na2SO4:
[tex]\begin{gathered} m\text{ = }\frac{m(Na2SO4)}{M(Na2SO4)}\times\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{3mol\text{ Na2SO4}}\times M(Fe2(SO4)3) \\ \text{ =}\frac{5}{142.04}\times\frac{1}{3}\times399.88 \\ \text{ = 4.69g} \end{gathered}[/tex]
Answer:
Therefore, FePO4 is the excess reactant and Na2SO4 is the limiting reactant
