$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$
$1$ mole of $B a C l_{2}$ and $2$ mole of $N a_{3} (P O_{4})_{2}$ are allowed to react together. The number of oxygen atoms that are present is;

32 \times 10^{23}

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24 \times 10^{23}

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12 \times 10^{23}

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48 \times 10^{23}

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Solution

The correct option is D $48 \times 10^{23}$
The balanced chemical equation is as given below:
$3 B a C l_{2} + 2 N a_{3} (P O_{4})_{2} \to B a_{3} (P O_{4})_{2} + 6 N a C l$
$3$ moles of $B a C l_{2}$ reacts with $2$ moles of $N a_{3} P O_{4}$ to produce $1$ mole of $B a_{3} (P O_{4})_{2}$ .
$1$ mole of $B a C l_{2}$ reacts with $0.67$ moles of $N a_{3} P O_{4}$ to produce $0.33$ mole of $B a_{3} (P O_{4})_{2}$ .
Thus, $1.33$ moles of $N a_{3} P O_{4}$ and $0.33$ moles of $B a_{3} (P O_{4})_{2}$ are present.
Hence, the number of moles of oxygen atoms present is $1.33 \times 4 + 0.33 \times 8 = 8$ moles.
The number of molecules of oxygen atom is $8 \times 6.023 \times 10^{23} = 48 \times 10^{23}$ .

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3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl 1 mole of BaCl2 and 2 mole of Na3(PO4)2 are allowed to react together. The number of oxygen atoms that are present is;

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