if i perform this reaction with 25 grams of iron III phosphate and an excess of sodium sulfate, how many grams of iron III sulfate can i make

What reaction? And use caps and parentheses when required.

To determine the number of grams of iron III sulfate that can be formed, you need to calculate the theoretical yield of the reaction. This can be done by balancing the chemical equation, converting the given mass of the reactant to moles, determining the limiting reactant, and then converting the moles of the limiting reactant to grams of the product.

1. Balance the chemical equation:
FePO4 + 3Na2SO4 → Fe2(SO4)3 + 2Na3PO4

2. Convert the given mass of iron III phosphate (FePO4) to moles using its molar mass. The molar mass of FePO4 is calculated by adding the atomic masses of its constituents: Fe (55.845 g/mol) + P (30.974 g/mol) + 4O (4 * 16.00 g/mol). Therefore, the molar mass of FePO4 is 150.827 g/mol.
Moles of FePO4 = Given mass (25 g) / Molar mass (150.827 g/mol)

3. Determine the limiting reactant by comparing the stoichiometric ratios of the reactants. Since there is an excess of sodium sulfate (Na2SO4), it will not limit the reaction. The stoichiometric ratio between FePO4 and Fe2(SO4)3 is 1:1. So, the moles of FePO4 are equal to the moles of Fe2(SO4)3 that can be formed.

4. Convert the moles of Fe2(SO4)3 to grams by multiplying with its molar mass. The molar mass of Fe2(SO4)3 is calculated by adding the atomic masses of its constituents: Fe (2 * 55.845 g/mol) + S (3 * 32.06 g/mol) + 12O (12 * 16.00 g/mol). Therefore, the molar mass of Fe2(SO4)3 is 399.881 g/mol.
Grams of Fe2(SO4)3 = Moles of Fe2(SO4)3 * Molar mass (399.881 g/mol)

By following these steps, you can calculate the number of grams of iron III sulfate that can be made from the given quantity of iron III phosphate.