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  4. If 0.5 moles of BaCl 2 are mixed with 0.2 moles of Na 3 PO 4 the maximum number of moles of Ba 3( PO 4)2 that can be formed, is

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Q. If $0.5$ moles of $BaCl _{2}$ are mixed with $0.2$ moles of $Na _{3} PO _{4}$ the maximum number of moles of $Ba _{3}\left( PO _{4}\right)_{2}$ that can be formed, is

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Solution:

$3 BaCl _{2}+2 Na _{3} PO _{4} \to Ba _{3}\left( PO _{4}\right)_{2}+6 NaCl$

Limiting reactant is $Na _{3} PO _{4}$

$0.2 \,mol\, Na _{3} PO _{4}$ will give $Ba _{3}\left( PO _{4}\right)_{2}$

$=\frac{1}{2} \times 0.2=0.1 \,mol$