Final answer:
To calculate the mass of copper (ii) phosphate required to produce 8.7 moles of aluminum phosphate, we need the balanced reaction and the molar mass of copper (ii) phosphate. This mass can then be calculated by multiplying the number of moles by the molar mass of copper (ii) phosphate.
Step-by-step explanation:
To determine the mass of copper (ii) phosphate required to produce 8.7 moles of aluminum phosphate, we need to understand the stoichiometry of the reaction. First and foremost, we need the balanced chemical equation of the reaction involving copper (ii) phosphate and aluminum. Unfortunately, without this balanced equation, we cannot proceed with certainty. However, if we assume a common reaction where aluminum displaces copper from copper (ii) phosphate to form aluminum phosphate, a hypothetical balanced equation might be:
3 Cu3(PO4)2 + 2 Al → 2 AlPO4 + 3 Cu
This is simply an illustration and not the actual equation. Nonetheless, once we have the correct balanced equation, we would use the molar mass of copper (ii) phosphate and stoichiometry to calculate the mass required.
For example, if the balanced equation dictated that 1 mole of copper (ii) phosphate produced 1 mole of aluminum phosphate, we would find the molar mass of copper (ii) phosphate using the molecular formula, multiply it by the number of moles required (8.7 in this case) to get the total mass.