9.3 g of phosphorous was taken in the reaction shown:

$P + 5 H N O_{3} ⟶ H_{3} P O_{4} + H_{2} O + 5 N O_{2}$

Calculate the volume of $N O_{2}$ produced at S.T.P..

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

33.6 L

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45 L

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50 L

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55 L

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Solution

The correct option is B 33.6 L
Given that,

Mass of P used, w = 9.3 g
Molar mass of P = 31 g/mol

Now,
Moles of phosphorous = $\frac{w}{M} = \frac{9.3}{31} = 0.3$ mol

According to reaction,

1 mol of P will produce 5 moles $N O_{2}$ .

So, volume of $N O_{2}$ produced $= 0.3 \times 5 \times 22.4 = 33.6 L$

Hence, option $A$ is correct.

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Similar questions

9.3 g of phosphorous was used in the reaction shown.

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

Calculate the number of moles of phosphorus taken.

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]

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]

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

The mass of phosphoric acid that can be prepared from 6.2 g of phosphorous and the mass of

H N O_{3}

that will be consumed at the same time is respectively :

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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9.3 g of phosphorous was taken in the reaction shown: P+5HNO3⟶H3PO4+H2O+5NO2Calculate the volume of NO2 produced at S.T.P..[H = 1, N = 14, P = 31, O = 16]

The correct option is B 33.6 LGiven that,Mass of P used, w = 9.3 gMolar mass of P = 31 g/molNow,Moles of phosphorous = wM=9.331=0.3 molAccording to reaction, 1 mol of P will produce 5 moles NO2.So, volume of NO2 produced =0.3×5×22.4=33.6 LHence, option A is correct.

9.3 g of phosphorous was taken in the reaction shown:

$P + 5 H N O_{3} ⟶ H_{3} P O_{4} + H_{2} O + 5 N O_{2}$

Calculate the volume of $N O_{2}$ produced at S.T.P..

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