If 0.5 mol of `CaBr_(2)` is mixed with 0.2 mol of `K_(3) PO_(4)`, the maximum nubmer of moles of `Ca_(3) (PO_(4))_(2)` that can be formed is:
a. 0.1 b. 0.2 c. 0.5 d.0.7

Explanation

To determine the maximum number of moles of \( \text{Ca}_3(\text{PO}_4)_2 \) that can be formed from the reaction between \( \text{CaBr}_2 \) and \( \text{K}_3\text{PO}_4 \), we first need to write the balanced chemical equation for the reaction.

### Step 1: Write the balanced chemical equation
The reaction can be represented as follows:

\[ 3 \text{CaBr}_2 + 2 \text{K}_3\text{PO}_4 \rightarrow 6 \text{KBr} + \text{Ca}_3(\text{PO}_4)_2 \]
...