Phosphoric acid H3 PO4 is prepared in a two-step process-1 P4 - 5O2 - P4 O102 P4 O10 - 6H2 O -4H3 PO4In step 1- 62 g of phosphorus is made to react with excess oxygen to form P4 O10 with 85

P_4 + 5O_2 → P_4 O_10 ….(1) P_4 O_10 + 6H_2 O→ nbsp;4H_3 PO_4 ….(2) Moles of P_4 = 62124 = 0.5 mol In reaction 1, 1 mol of P_4 produces 1 mol of P_4 O_10 0.5 ...

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Standard XIII

Chemistry

Stoichiometric Calculations

Phosphoric ac...

Question

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.

375 g

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150 g

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125 g

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264 g

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Solution

The correct option is B $150 g$
$P_{4} + 5 O_{2} \to P_{4} O_{10} \dots . (1)$
$P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4} \dots . (2)$
Moles of $P_{4} = \frac{62}{124} = 0.5 mol$
In reaction 1,
1 mol of $P_{4}$ produces 1 mol of $P_{4} O_{10}$
0.5 mol of $P_{4}$ will produce 0.5 mol of $P_{4} O_{10}$
Yield of the reaction is given as 85 %
Moles of $P_{4} O_{10}$ formed $= 0.85 \times 0.5 = 0.425$ mol
In reaction 2,
1 mol of $P_{4} O_{10}$ produces 4 mol of $H_{3} P O_{4}$
0.425 mol will produce $(4 \times 0.425) = 1.7$ mol of $H_{3} P O_{4}$
Yield of the reaction is given as 90 %
Moles of $H_{3} P O_{4}$ formed $= 1.7 \times 0.9 = 1.53 mol$
Mass of $H_{3} P O_{4}$ formed $= 98 \times 1.53 = 150 g$

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Similar questions

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} {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:

Q. Phosphoric acid (

H_{3} P O_{4}

) is widely used to make fertilizers and can be prepared in a two-step process.

S t e p I : P_{4} + 5 O_{2} ⟶ P_{4} O_{10}

S t e p I I : P_{4} O_{10} + 6 H_{2} O ⟶ 4 H_{3} P O_{4}

We allow

310

g of phosphorus to react with excess oxygen, which forms tetraphosphorus deoxide (

P_{4} O_{10}

) in

50 %

yield. In the step

I I

25 %

yield of

H_{3} P O_{4}

is obtained.
(Atomic weight in amu of

P = 31, H = 1, O = 16

)

Assuming

100 %

yield, the number of moles of

H_{3} P O_{4}

is :

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)

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Phosphoric acid (H3PO4) is prepared in a two-step process: (1) P4+5O2→P4O10 (2) P4O10+6H2O→4H3PO4 In step (1), 62 g of phosphorus is made to react with excess oxygen to form P4O10 with 85 % yield. In step (2), the yield of H3PO4 obtained is 90 %. Calculate the mass of H3PO4 produced.