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If 0.5 mol of CaBr(2) is mixed with 0.2 ...

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

A

0.1

B

0.2

C

0.5

D

0.7

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(2 "mol")(3 BaCl_(2)) + underset(2 "mol")(2Na_(3)PO_(4)) rarr underset(6 "mol")(6NaCl) + underset(1 "mol")(Ba_(3) (PO_(4))_(2))`
Given `implies 0.5 "mol of" BaCl_(2)` and `0.2 "mol of" Na_(3)PO_(4)`
Find the limiting reagent:
`2 "mol of" Na_(3)PO_(4) implies 2 "mol of" BaCl_(2)`
`0.2 "mol of" Na_(3)PO_(4) implies 0.3 "mol of" BaCl_(2)`
`:. Na_(3)PO_(4)` is the limiting reagent
`:. "mol of" Na_(3)PO_(4) implies "mol of" Ba_(3) (PO_(4))_(2)`
`0.2 "mol" of NaPO_(4) implies 0.1 "mol of" Ba_(3)(PO_(4))_(2)`
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