Final answer:
To balance the reaction of tin with iron(II) phosphate to form tin(IV) phosphate and iron, the balanced chemical equation is 3Sn + 2Fe3(PO4)2 → Sn3(PO4)4 + 6Fe.
Step-by-step explanation:
To write a balanced chemical equation for the reaction where tin reacts with iron(II) phosphate to form tin(IV) phosphate and iron precipitate, we first need to write down the correct formulas for the reactants and products. We then balance the equation to ensure that the number of atoms for each element on the reactant side is equal to the number on the product side.
The reactants are tin (Sn) and iron(II) phosphate (Fe3(PO4)2), and the products are tin(IV) phosphate (Sn3(PO4)4) and iron (Fe). Writing down the unbalanced equation, we get:
Sn + Fe3(PO4)2 → Sn3(PO4)4 + Fe
To balance the equation, notice that we have 3 iron(II) ions on the reactant side and just 1 iron atom on the product side. Accordingly, we need three iron atoms on the product side to balance it. Similarly, there is one tin atom on the reactant side and three in the product as tin(IV) phosphate. Hence, we need to multiply the number of tin atoms on the reactant side by three. Upon balancing the phosphates and iron, we get the final balanced chemical equation:
3Sn + 2Fe3(PO4)2 → Sn3(PO4)4 + 6Fe