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Haber Bosch Process

If 0.5 mole...

Question

If $0.5$ mole of $B a C l_{2}$ is mixed with $0.20$ mole of $N a_{3} P O_{4}$ , the maximum number of $B a_{3} (P O_{4})_{2}$ that can be formed is:

A

0.70

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B

0.50

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C

0.20

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D

0.10

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Solution

The correct option is D $0.10$
$3 {B a C l}_{2} + 2 {N a}_{3} {P O}_{4} ⟶ {B a}_{3} {({P O}_{4})}_{2} + 6 N a C l$

$3$ mol of ${B a C l}_{2}$ reacts with $2$ mole of ${N a}_{3} {P O}_{4}$ to give $1$ mole of ${B a}_{3} {({P O}_{4})}_{2}$ and $6$ mole of NaCl.

$∴$ 0.5 mol $B a C l_{2}$ reacts wtih 0.33 mol of $N a_{3} P O_{4}$ .

Here, ${N a}_{3} {P O}_{4}$ is the limiting reagent.

$2 m o l e s o f {N a}_{3} {P O}_{4} ⟶ 1 m o l e {B a}_{3} {({P O}_{4})}_{2}$

$1 m o l e o f {N a}_{3} {P O}_{4} ⟶ \frac{1}{2} m o l e {B a}_{3} {({P O}_{4})}_{2}$

$0.2 m o l e o f {N a}_{3} {P O}_{4} ⟶ \frac{1}{2} \times 0.2 m o l e {B a}_{3} {({P O}_{4})}_{2}$

$⟶ 0.1 m o l e {B a}_{3} {({P O}_{4})}_{2}$

The maximum number of mole of ${B a}_{3} {({P O}_{4})}_{2}$ is $0.1$ mole.

Hence, option $D$ is correct.

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Q. If 0.5 moles of

B a C l_{2}

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If 0.5 mol of $B a C l_{2}$ is mixed with 0.2 mol 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:

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