Answer: The theoretical yield of iron(III) sulfate is 26.6 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of iron(III) phosphate = 20.00 g
Molar mass of iron(III) phosphate = 150.82 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol[/tex]
The given chemical equation follows:
[tex]2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4[/tex]
As, sodium sulfate is present in excess. So, it is considered as an excess reagent.
Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate
So, 0.133 moles of iron(III) phosphate will produce = [tex]\frac{1}{2}\times 0.133=0.0665moles[/tex] of iron(III) sulfate
Now, calculating the mass of iron(III) sulfate from equation 1, we get:
Molar mass of iron(III) sulfate = 399.9 g/mol
Moles of iron(III) sulfate = 0.0665 moles
Putting values in equation 1, we get:
[tex]0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g[/tex]
Hence, the theoretical yield of iron(III) sulfate is 26.6 grams
