what are the products of the reaction?
Cu3PO4 + HCl =
To determine the products of the reaction between Cu3PO4 (copper(I) phosphate) and HCl (hydrochloric acid), we need to consider the general rules of acid-base and double displacement reactions.
In this case, HCl is an acid and Cu3PO4 is a base (or metal phosphate). When an acid reacts with a base, they usually undergo a neutralization reaction, resulting in the formation of a salt and water.
The reaction can be written as follows:
Cu3PO4 + 3HCl → 3CuCl2 + H3PO4
In this reaction, copper(I) phosphate reacts with hydrochloric acid to produce copper(II) chloride and phosphoric acid.
It is important to note that the subscripts must be balanced, indicating the number of atoms or ions present in the compounds. In this case, Cu3PO4 becomes 3CuCl2 since copper(I) ions combine with two chloride ions to form copper(II) chloride. Additionally, the balanced equation shows that three moles of hydrochloric acid react with one mole of copper(I) phosphate.