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

`underset(2 mol)(3CaBr_(2)) + underset(2 mol)(2K_(3) PO_(4)) rarr underset(6 mol)(6 KBr) + underset(1 mol)(Ca_(3) (PO_(4))_(2))`
According to the chemcial reaction, 2 mol of `K_(3) PO_(4)` reacts = 3 mol of `CaBr_(2)`
`0.2 mol of K_(3) PO_(4)` reacts `= 0.3 mol of CaBr_(2)`
Moles of `CaBr_(2)` left unreacted `= 0.5 - 0.3 = 0.2`
Hence, `K_(3) PO_(4)` is the limiting reagent, since it reacts completely and it deiceds the amount of products, thus formed
2 mol of `K_(3) PO_(4)` gives `= 1 mol of Ca_(3) (PO_(4))_(2)`
0.2 mol of `K_(3) PO_(4)` gives `= 0.1 mol` of `Ca_(3) (PO_(4))_(2)`
Thus, the answers is (a)