Question

What is the limiting and excess reactant if 15.0g of FePo4 reacts with 5.0g of Na2SO

Answer 1

Step-by-step 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

`\begin{gathered} m\text{ = }(m(FePO4))/(M(FePO4))*\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{2mol\text{ FePO4}}*\text{ M\lparen Fe2\lparen SO4\rparen3\rparen} \\ \text{ = }(15)/(150.82)*(1)/(2)*399.88 \\ \text{ = 19.89g} \end{gathered}`

We then find the mass of Fe2(SO4)3 from Na2SO4:

`\begin{gathered} m\text{ = }(m(Na2SO4))/(M(Na2SO4))*\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{3mol\text{ Na2SO4}}* M(Fe2(SO4)3) \\ \text{ =}(5)/(142.04)*(1)/(3)*399.88 \\ \text{ = 4.69g} \end{gathered}`

Answer:

Therefore, FePO4 is the excess reactant and Na2SO4 is the limiting reactant