If 15.0 grams of Iron (III) oxide reacts with 25.0 grams of Strontium phosphate, how many grans of Iron (III) phosphate will be produced?
Probably none. Fe2O3 is not soluble. Sr3(PO4)2 is insoluble. No reaction.
To determine the number of grams of Iron (III) phosphate produced, we need to calculate the moles of reactants and use the balanced chemical equation to find the stoichiometric ratio between Iron (III) oxide and Iron (III) phosphate.
1. Start by finding the moles of Iron (III) oxide (Fe2O3):
- Use the molar mass of Fe2O3 to convert grams to moles.
- The molar mass of Iron (III) oxide is calculated as follows:
- Atomic mass of Fe: 55.85 g/mol
- Atomic mass of O: 16.00 g/mol (there are three O atoms)
- Total molar mass of Fe2O3: (2 * 55.85) + (3 * 16.00) = 159.69 g/mol
- Moles of Fe2O3 = (mass of Fe2O3) / (molar mass of Fe2O3)
- Moles of Fe2O3 = 15.0 g / 159.69 g/mol
2. Next, find the moles of Strontium phosphate (Sr3(PO4)2):
- Use the molar mass of Sr3(PO4)2 to convert grams to moles.
- The molar mass of Strontium phosphate is calculated as follows:
- Atomic mass of Sr: 87.62 g/mol (there are three Sr atoms)
- Atomic mass of P: 30.97 g/mol (there are two P atoms)
- Atomic mass of O: 16.00 g/mol (there are eight O atoms)
- Total molar mass of Sr3(PO4)2: (3 * 87.62) + (2 * 30.97) + (8 * 16.00) = 452.38 g/mol
- Moles of Sr3(PO4)2 = (mass of Sr3(PO4)2) / (molar mass of Sr3(PO4)2)
- Moles of Sr3(PO4)2 = 25.0 g / 452.38 g/mol
3. Determine the limiting reactant:
- Compare the moles of Fe2O3 and Sr3(PO4)2 from steps 1 and 2.
- The reactant with lower moles is the limiting reactant because it determines the maximum amount of product.
- In this case, whichever reactant has fewer moles is the limiting reactant.
4. Use the balanced chemical equation to find the stoichiometric ratio:
- The balanced chemical equation for the reaction is as follows:
2 Fe2O3 + 3 Sr3(PO4)2 -> 6 FePO4 + Sr6(PO4)2
- According to the balanced equation, each 2 moles of Fe2O3 react with 3 moles of Sr3(PO4)2 to produce 6 moles of FePO4.
- This corresponds to the stoichiometric ratio of:
2 moles Fe2O3 : 6 moles FePO4
5. Calculate the moles of Iron (III) phosphate produced:
- Since the limiting reactant determines the maximum yield of product, we can use the stoichiometric ratio from step 4 to find the moles of FePO4 produced.
- Moles of FePO4 = (moles of Fe2O3) * (6 moles FePO4 / 2 moles Fe2O3)
6. Convert the moles of Iron (III) phosphate to grams:
- Use the molar mass of FePO4 to convert moles to grams.
- The molar mass of Iron (III) phosphate is calculated as follows:
- Atomic mass of Fe: 55.85 g/mol
- Atomic mass of P: 30.97 g/mol
- Atomic mass of O: 16.00 g/mol (there are four O atoms)
- Total molar mass of FePO4: 55.85 + 30.97 + (4 * 16.00) = 150.82 g/mol
- Mass of FePO4 = (moles of FePO4) * (molar mass of FePO4)
By following these steps, you can calculate the number of grams of Iron (III) phosphate produced in the reaction between 15.0 grams of Iron (III) oxide and 25.0 grams of Strontium phosphate.