Concentrate nitric acid oxidises phosphorus to phosphoric acid according to the following equation- P - 5HNO3conc- H3PO4 - H2O - 5NO2If 9-3 g of phosphorus was used in the reaction- calculate-The volume of nitrogen dioxide produced at STP- -H - 1- N - 14- P - 31- O - 16-

P + 5HNO_3 → H_3PO_4 + H_2O + 5NO_2 5NO_2 = #160; 5× 22.4 = 112 𝑙 .Molecular weight of H_3PO_4 = 3+31+ 16× 4 = 98g.If 31g of ph ...

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Standard X

Chemistry

Mole Concept

Concentrate n...

Question

Concentrate nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
$P + 5 H N O_{3} (c o n c .) \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}$
If $9.3 g$ of phosphorus was used in the reaction, calculate:
The volume of nitrogen dioxide produced at $S T P$ .
$[H = 1, N = 14, P = 31, O = 16]$

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Solution

$P + 5 H N O_{3} \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}$
$5 N O_{2}$ = $5 \times 22.4$ = 112 $l$ .
Molecular weight of $H_{3} P O_{4}$ = 3+31+ $16 \times 4$ = 98g.
If 31g of phosphorus releases 112 $l$ of $H_{3} P O_{4}$
Then, 9.3g of phosphorus releases = $\frac{112}{31} \times 9.3$ = 33.6 $l$ .

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Q. Concentrate nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5 H N O_{3} (c o n c .) \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}

9.3 g

of phosphorus was used in the reaction, calculate:
The mass of phosphoric acid formed.

Q. Concentrate nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5 H N O_{3} (c o n c .) \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}

What mass of phosphoric acid can be prepared from

6.2 g

of phosphorous?

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
$P + 5 H N O_{3} \to H_{3} P O_{4} + 5 N O_{2} + H_{2} O$
If 6.2 g of phosphorus was used in the reaction, calculate:

(a) The number of moles of phosphorus taken and mass of phosphoric acid formed.
(b) Mass of nitric acid consumed at the same time.
(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and $273^{\circ} C .$

(a) Concentrated nitric acid oxidises phosphorus to phosphhoric acid according to the following equation;
$P + 5 H N O_{3} (c o n c .) \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}$
If 9.3 g of phosphorus was used in the reaction, calculate;
(i) Number of mles of phosphorus take.
(ii) The mass of phosphoric acid formed.
(iii) The volume of nitrogen dioxide produced at STP.
(b) (i) 67.2 Litres of hydrogen combines with 44.8 letres of nitrogen to form ammonia under specific conditions as:
N_2(g) +3H_2(g) \rightarrow 2NH_3(g)
Calculate the volume of ammonia produced.
What is the other substance, if any that remains in the resultant mixture?
(ii) The mass of $5.6 d m^{3}$ of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.
(iii) Find the total percentage of Magnesium in magnisum nitrate crystals, $M g (N O_{3})_{2} . 6 H_{2} O .$
$[M g = 24, N = 14, O = 16 a n d H = 1]$

P + 5 H N O_{3} \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}

What would be the volume of steam produced from 6.2 g of phosphorus at

S T P

[H = 1; N = 14; O = 16; P = 31]

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Concentrate nitric acid oxidises phosphorus to phosphoric acid according to the following equation:P+5HNO3(conc.)→H3PO4+H2O+5NO2If 9.3 g of phosphorus was used in the reaction, calculate:The volume of nitrogen dioxide produced at STP.[H=1,N=14,P=31,O=16]

P+5HNO3→H3PO4+H2O+5NO25NO2 = 5×22.4 = 112l.Molecular weight of H3PO4 = 3+31+16×4 = 98g.If 31g of phosphorus releases 112l of H3PO4Then, 9.3g of phosphorus releases = 11231×9.3 = 33.6l.

$P + 5 H N O_{3} \to H_{3} P O_{4} + H_{2} O + 5 N O_{2}$
$5 N O_{2}$ = $5 \times 22.4$ = 112 $l$ .
Molecular weight of $H_{3} P O_{4}$ = 3+31+ $16 \times 4$ = 98g.
If 31g of phosphorus releases 112 $l$ of $H_{3} P O_{4}$
Then, 9.3g of phosphorus releases = $\frac{112}{31} \times 9.3$ = 33.6 $l$ .