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Standard XII

Chemistry

Laws of Mass Conservation

Phosphoric ac...

Question

Phosphoric acid is prepared in a two-step process. Find the mass of $H_{3} P O_{4}$ formed when 124 g of $P_{4}$ reacts with an excess of oxygen and then treated with sufficient water.
$P_{4} + O_{2} ⟶ P_{4} O_{10}$
$P_{4} O_{10} + H_{2} O ⟶ H_{3} P O_{4}$
(Molar mass of P = 31 g/mol)

150 g

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288 g

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560 g

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392 g

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Solution

The correct option is D 392 g
Let the weight of acid produced be $X g$ .
Since the atoms of $P$ are conserved,
Applying POAC on phosphorus,
$4 \times moles of P_{4} = 1 \times moles of H_{3} P O_{4}$

$\Rightarrow 4 \times \frac{weight of P_{4}}{molar mass of P_{4}} = 1 \times \frac{weight of acid}{molar mass of acid}$

$\Rightarrow 4 \times \frac{124}{124} = 1 \times \frac{X}{98}$

On solving, $X = 392 g$
Therefore,the mass of phosphoric acid produced is 392 g.

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Similar questions

Q. Phosphoric acid is prepared in a two-step process. Find the mass of

H_{3} P O_{4}

formed when 124 g of

P_{4}

reacts with an excess of oxygen and then treated with sufficient water.

P_{4} + O_{2} ⟶ P_{4} O_{10}

P_{4} O_{10} + H_{2} O ⟶ H_{3} P O_{4}

(Molar mass of P = 31 g/mol)

Q. Phosphoric acid

(H_{3} P O_{4})

is prepared in a two-step process:
(1)

P_{4} + 5 O_{2} \to P_{4} O_{10}

(2)

P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}

In step (1),

62 g

of phosphorus is made to react with excess oxygen to form

P_{4} O_{10}

with 85 % yield. In step (2), the yield of

H_{3} P O_{4}

obtained is 90 %. Calculate the mass of

H_{3} P O_{4}

produced.

Q. Phosphoric acid (

H_{3} P O_{4}

) is prepared in a two-step process.

P_{4} + 5 O_{2} \to P_{4} O_{10}

P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}

186 g of phosphorus is made to react with excess oxygen which forms

P_{4} O_{10}

that has a 60 % yield. In the second reaction, 80% yield of

H_{3} P O_{4}

is obtained. The mass of

H_{3} P O_{4}

produced is:
(Molar mass of phosphorous = 31 g/mol)

Q. Phosphoric acid

(H_{3} P O_{4})

is prepared in a two-step process:
(1)

P_{4} + 5 O_{2} \to P_{4} O_{10}

(2)

P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}

In step (1),

62 g

of phosphorus is made to react with excess oxygen to form

P_{4} O_{10}

with 85 % yield. In step (2), the yield of

H_{3} P O_{4}

obtained is 90 %. Calculate the mass of

H_{3} P O_{4}

produced.

Q. Phosphoric acid

(H_{3} {PO}_{4})

prepared in a two step process.
(1)

P_{4} {+ 5O}_{2} \to P_{4} O_{10}

(2)

P_{4} O_{10} + {6H}_{2} O \to {4H}_{3} {PO}_{4}

We allow 62 g of phosphorus to react with excess oxygen which form

P_{4} O_{10}

which gives 85% yield. If in the step (2) reaction, 90% yield of

H_{3} {PO}_{4}

is obtained then mass of

H_{3} {PO}_{4}

produced is:

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Laws of Mass Conservation

Standard XII Chemistry

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Phosphoric acid is prepared in a two-step process. Find the mass of H3PO4 formed when 124 g of P4 reacts with an excess of oxygen and then treated with sufficient water. P4+O2⟶P4O10 P4O10+H2O⟶H3PO4 (Molar mass of P = 31 g/mol)

Phosphoric acid is prepared in a two-step process. Find the mass of $H_{3} P O_{4}$ formed when 124 g of $P_{4}$ reacts with an excess of oxygen and then treated with sufficient water.
$P_{4} + O_{2} ⟶ P_{4} O_{10}$
$P_{4} O_{10} + H_{2} O ⟶ H_{3} P O_{4}$
(Molar mass of P = 31 g/mol)