Fill in the coefficients that will balance the following reaction: a0PbI4 + a1CrCl3 → a2PbCl4 + a3CrI3
PbI₄ + CrCl₃ --> PbCl₄ + CrI₃
There is 1 Pb and Cr atom on both sides.
There are 3 Cl atoms on the left side and 4 on the right side.
There are 4 I atoms on the left side and 3 on the right side.
PbI₄ + 4CrCl₃ --> 3PbCl₄ + CrI₃
Cl atoms are now balanced
3PbI₄ + 4CrCl₃ --> 3PbCl₄ + 4CrI₃
All of the atoms are now balanced.
3PbI₄ + 4CrCl₃ --> 3PbCl₄ + 4CrI₃