If $0.5$ moles of $B a C l_{2}$ is mixed with $0.2$ moles of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed is:

0.7

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0.5

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0.03

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0.10

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Solution

The correct option is D $0.10$
The reaction is $B a C l_{2} + 2 {N a}_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$ .

$1$ mole of $B a C l_{2}$ reacts with $2$ moles of $N a_{3} P O_{4}$ to form $1$ mole of $B a_{3} (P O_{4})_{2}$ .

So, we can see that sodium phosphate is the limiting reagent.

$0.1$ moles of $B a C l_{2}$ will react with $0.2$ moles of $N a_{3} P O_{4}$ to form $0.10$ moles of $B a_{3} (P O_{4})_{2}$ .

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Q. 0.5 mole

B a C l_{2}

is mixed with 0.2 mole

N a_{3} P O_{4}

, the maximum number of moles of

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Q. If 0.5 moles of

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is mixed with 0.2 moles of

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If $0.5 m o l e$ of $B a C l_{2}$ is mixed with $0.2 m o l e$ of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ formed are

Q. If 0.5 moles of

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If 0.5 moles of BaCl2 is mixed with 0.2 moles of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is:

The correct option is D 0.10The reaction is BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl.1 mole of BaCl2 reacts with 2 moles of Na3PO4 to form 1 mole of Ba3(PO4)2.So, we can see that sodium phosphate is the limiting reagent. 0.1 moles of BaCl2 will react with 0.2 moles of Na3PO4 to form 0.10 moles of Ba3(PO4)2.

If $0.5$ moles of $B a C l_{2}$ is mixed with $0.2$ moles of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed is: