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Standard XII

Chemistry

Hybridisation in Carbon

One mole, of ...

Question

One mole, of $P_{4} O_{10}$ is allowed to react fully with dust and salt-free doubly distilled water and the volume is made up to 1L. What is the normality of the so-generated orthophosphoric acid?
$P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}$

1.0 N

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8.0 N

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12.0 n

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4.0 N

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Solution

The correct option is A 12.0 n
$P_{4} O_{10} 1 m o l + 6 H_{2} O \to 4 H_{3} P O_{4} 4 m o l$
Thus, 4 moles of orthophosphoric acid are generated in 1L solution.
$∴$ Mobility = 4M a
nd normality = molarity $\times$ basicity $= 4 \times 3 = 12.0 N$

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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 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. What weight of each compound will be obtained when

0.450

g of

P_{4} O_{10}

and 1.50 g of

P C l_{5}

are allowed to react completely?
Given reaction:

6 P C l_{5} + P_{4} O_{10} \to 10 P O C l_{3}

Q. What weight of each compound will be obtained when

0.450

g of

P_{4} O_{10}

and 1.50 g of

P C l_{5}

are allowed to react completely?
Given reaction:

6 P C l_{5} + P_{4} O_{10} \to 10 P O C l_{3}

Q. Two substance,

P_{4}

and

O_{2}

are allowed to react completely to form mixture of

P_{4} O_{6}

and

P_{4} O_{10}

leaving none of the reactants. The composition of final mixture if

P_{4}

and

O_{2}

is taken as:

P_{4} + 3 O_{2} \to P_{4} O_{6}

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

(i) If

1

mole

P_{4}

;

4

moles of

O_{2}

(ii) If

3

moles

P_{4}

;

11

moles of

O_{2}

(iii) If

3

moles

P_{4}

;

13

moles of

O_{2}

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One mole, of P4O10 is allowed to react fully with dust and salt-free doubly distilled water and the volume is made up to 1L. What is the normality of the so-generated orthophosphoric acid?P4O10+6H2O→4H3PO4

The correct option is A 12.0 nP4O101mol+6H2O→4H3PO44molThus, 4 moles of orthophosphoric acid are generated in 1L solution. ∴ Mobility = 4M and normality = molarity × basicity =4×3=12.0N

One mole, of $P_{4} O_{10}$ is allowed to react fully with dust and salt-free doubly distilled water and the volume is made up to 1L. What is the normality of the so-generated orthophosphoric acid?
$P_{4} O_{10} + 6 H_{2} O \to 4 H_{3} P O_{4}$

The correct option is A 12.0 n
$P_{4} O_{10} 1 m o l + 6 H_{2} O \to 4 H_{3} P O_{4} 4 m o l$
Thus, 4 moles of orthophosphoric acid are generated in 1L solution.
$∴$ Mobility = 4M a
nd normality = molarity $\times$ basicity $= 4 \times 3 = 12.0 N$