Question

Calculate the number of moles of Ba3(PO4)2 that can be formed if 0.5 mole of BaCl2 is mixed with 0.2 mole of Na3PO4.​

Answer 1

Answer:

$0.1$ mole

Explanation:

The reaction will be:

$3BaCl_{2} +2Na_{3}PO_{4}$ → $Ba_{3} (PO_{4})_{2} +6NaCl$

We can find out the ratio of the reactants and products with the help of the coefficients of the balanced equation written above.

• The required ratio of $BaCl_{2} : Na_{3}PO_{4}$ is $3:2$
• $0.5$ mole $BaCl_{2}$ requires $\frac{2}{3} \times 0.5$ mole $Na_{3}PO_{4}$. You have only $0.2$ mole $Na_{3}PO_{4}$ so it is limiting and $BaCl_{2}$ is in excess.
• $2$ moles of $Na_{3}PO_{4}$ will yield $1$ mole of $Ba_{3}(PO_{4} )_{2}$.
• $0.2$ moles $Na_{3}PO_{4}$ will yield $0.1$ mol of $Ba_{3}(PO_{4} )_{2}$.

Hence, $0.1$ moles of $Ba_{3}(PO_{4} )_{2}$ will be formed.

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Answer 2

Answer:

0.1 moles of $Ba_{3} (PO_{4})_{2}$ will be formed.

Explanation:

The balanced equation for this reaction is:

$3BaCl_{2} + 2Na_{3} PO4 --- > Ba_{3} (PO_{4} )_{2} + 6NaCl$

Given moles of BaCl2 are 0.5

Given moles of Na3PO4 are 0.2

To find out the Limiting Reagent:

3 moles of BaCl2 require 2 moles of Na3PO4

1 mole of BaCl2 requires 2/3 moles of Na3PO4

0.5 moles of BaCl2 require 0.33 moles of Na3PO4

But, according to question we have only 0.2 moles of Na3PO4. Therefore, Na3PO4 is the L.R. and will govern the reaction.

So, 2 moles of Na3PO4 give 1 mole of Barium Phosphate.

1 mole of Na3PO4 gives 0.5 moles of Barium Phosphate.

0.2 moles Sodium Phosphate will give 0.1 mole of Barium phosphate.

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