Question

CaBr2(aq)+2AuClO4(aq) → 2AuBr(s)+Ca(ClO4)2(aq)

The molecular equation you determined in Part B is shown above for your convenience. Examine each of the chemical species involved to determine the ions that would be present in solution.

Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation.

Answer 1

To write a complete ionic equation, I would first separate each of the elements and their respective charges, subscripts, and coefficients.

For CaBr2(aq):

Ca has a 2+ charge and a coefficient of 1. Ca+

Br has a charge of -1 and has a subscript of 2. Br-

For 2AuClO4(aq):

Au has a charge of 1+ because perchlorate (ClO4) has a charge of -1, so they must be "cancel" in charges in order to form. Au also has a coefficient of two. 2Au+

ClO4 has a charge of -1 and has a coefficient of 2. 2ClO4-

For 2AuBr(s):

You can leave this alone in the complete ionic equation since it is a solid precipitate of the reaction and has little to no ions dissociating.

For Ca(ClO4)2(aq):

Ca has a charge of 2+ and a coefficient of 1. Ca2+

ClO4 has a charge of 1-, a coefficient of one, but also a subscript of 2. 2ClO4-

Thus, when separating and combining everything, you should be left with:

`Ca^(2+)(aq) + 2Br^(-)(aq) + 2Au^(+)(aq) + 2ClO4^(-)(aq) --- > 2AuBr(s) + Ca^(2+)(aq) + 2ClO4^(-)(aq)`

Hope this helps!