Question

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

• A. 0.7
• B. 0.5
• C. 0.30
• D. 0.10

Answer 1

Answer: (D)

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

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

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

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