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Standard XII

Chemistry

Sulphur

Phosphoric ac...

Question

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 :

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Solution

$310 g m P_{4} = \frac{310}{124} m o l P_{4} = 2.5 m o l P_{4} = 2.5 \times 4 = 10 m o l H_{3} P O_{4}$

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Similar questions

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:

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. In a particular experiment,

310 g

of phosphorous (

P_{4}

) reacted with excess of oxygen to form

P_{4} O_{10}

in 75% yield. Then

P_{4} O_{10}

formed reacted with

54

g of water to give

H_{3} P O_{4} .

Amount of

H_{3} P O_{4}

obtained (in gm) is :
(Consider only two places after decimal for calculation)

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310gmP4=310124molP4=2.5molP4=2.5×4=10molH3PO4

$310 g m P_{4} = \frac{310}{124} m o l P_{4} = 2.5 m o l P_{4} = 2.5 \times 4 = 10 m o l H_{3} P O_{4}$