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Stoichiometry

3BaCl2+2Na2PO...

Question

$3 B a C l_{2} + 2 N a_{2} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$
Maximum amount of $B a_{3} (P O_{4})_{2}$ formed when $2$ moles each of $N a_{3} P O_{4}$ and $B a C l_{2}$ react is:

4 m o l

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1 m o l

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\frac{2}{3} m o l

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\frac{5}{3} m o l

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Solution

The correct option is A $\frac{5}{3} m o l$
$3 B a C l_{2} + 2 N a_{3} P O_{4} ⟶ B a_{3} (P O_{4})_{2} + 6 N a C l$
$3$ $m o l e s$ of $B a C l_{2}$ gives $1$ $m o l e$ of $B a_{3} (P O_{4})_{2}$
$∴$ $2$ $m o l e s$ of $B a C l$ give $\frac{2}{3}$ $m o l e s$ of $B a_{3} (P O_{4})_{2}$
$2$ $m o l e$ of $N a_{3} P O_{4}$ gives $1$ $m o l e$ of $B a_{3} (P O_{4})_{2}$
$∴$ $2$ $m o l e s$ of $N a_{3} P O_{4} ⟶$ $1$ $m o l e$ of $B a_{3} (P O_{4})_{2}$
Total moles of $B a_{3} (P O_{4})_{3} = 1 + \frac{2}{3} = \frac{5}{3}$ $m o l e s$

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B a C l_{2}

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Calculate n-factor of $N a_{3} P O_{4}$ in the following reaction.

$N a_{3} P O_{4}$ + $3 B a C l_{2}$ $\to$ 6NaCl + $B a_{3} (P O_{4})_{2}$

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