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 $273^{\circ} C .$

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Solution

$left parenthesis a right parenthesis 1 space m o l e space o f space p h o s p h o r u s space a t o m space equals 31 g space o f space p h o s p h o r u s space 31 g space o f space P space equals 1 space m o l e space o f space P 6.2 g space o f space P space equals fraction numerator 6.2 cross times 1 over denominator 31 end fraction equals 0.2 space m o l e s space o f space P left parenthesis b right parenthesis 31 g space P space r e a c t s space w i t h space H N O subscript 3 space equals 315 g s o space comma space 6.2 space g space P space w i l l space r e a c t space w i t h space H N O subscript 3 space equals 315 cross times fraction numerator 6.2 over denominator 31 end fraction equals 63 g left parenthesis c right parenthesis M o l e s space o f space s t e a m space f o r m e d space f r o m space 31 g space p h o s p h o r u s space equals 18 over 18 equals 1 m o l m o l e s space o f space s t e a m space f o r m e d space f r o m space 6.2 space g space p h o s p h o r u s space equals fraction numerator 6.2 over denominator 31 end fraction equals 0.2 space m o l V o l u m e space o f space s t e a m space p r o d u c e d space a t space S T P space equals 0.2 cross times 22.4 equals 4.48 l i t r e S i n c e space t h e space p r e s s u r e space left parenthesis 760 comma comma right parenthesis space r e m a i n s space c o n s tan t space comma b u t space t h e space t e m p e r a t u r e space left parenthesis 273 plus 273 right parenthesis space equals 546 space i s space d o u b l e space comma t h e space v o l u m e space o f space s t e a m space a l s o space g e t s space d o u b l e d space s o space comma space v o l u m e space o f space s t e a m space p r o d u c e d space a t space 760 m m space H g space a n d space 273 degree c space equals 4.48 cross times 2 equals 8.96 space l i t r e$

Volume is directly proportional to temperature at constant pressure(charles law).
(a) 1 mol of phosphorus forms 1 mol of phosphoric acid. Thus 0.2 mol of phosphorus will form 0.2 mol of phosphoric acid.
1 mol of phosphoric acid = (3 + 31 + 64) g = 98 g
0.2 mol of phosphoric acid = $98 \times 0.2$ = 49 g

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Similar questions

(a) Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
P + 5HNO₃ $\overset{}{\to}$ H₃PO₄ + H₂O + 5NO₂
(i) What mass of phosphoric acid can be prepared from 6.2 g of phosphorus?
(ii) What mass of nitric acid will be consumed at the same time?
(iii) What will be the volume of steam at the same time measured at 760 mm Hg pressure and 373°C? (H = 1; N = 14; O = 16; P = 31)

(b) Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation:
4NH₃ + 5O₂ $\overset{}{\to}$ 4NO + 6H₂O
If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

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?

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 volume of nitrogen dioxide produced at

S T P

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

(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, $M g (N O_{3})_{2} . 6 H_{2} O .$
$[M g = 24, N = 14, O = 16 a n d H = 1]$

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Volume is directly proportional to temperature at constant pressure(charles law). (a) 1 mol of phosphorus forms 1 mol of phosphoric acid. Thus 0.2 mol of phosphorus will form 0.2 mol of phosphoric acid. 1 mol of phosphoric acid = (3 + 31 + 64) g = 98 g 0.2 mol of phosphoric acid = 98×0.2 = 49 g