$4 N_{2} O + C H_{4} \to C O_{2} + 2 H_{2} O + 4 N_{2}$
If all volumes are measured at the same temperature and pressure, the volume of $N_{2} O$ required to give $150 c m^{3}$ of steam is :

200 cc

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150 cc

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420 cc

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300 cc

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Solution

The correct option is D 300 cc
Reaction:
$4 N_{2} O + C H_{4} \to C O_{2} + 2 H_{2} O + 4 N_{2}$
4 1 1 2 4
4 volumes of $N_{2} O$ will give 2 volumes of steam.
So, 300 cc of $N_{2} O$ will give 150 cc steam.

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Similar questions

The reaction 4N₂O + CH₄ $\overset{}{\to}$ CO₂ + 2H₂O + 4N₂ takes place in the gaseous state. If the volumes of all the gases are measured at the same temperature, and pressure, calculate the volume of dinitrogen oxide (N₂O), required to give 150 cm³ of steam.

(b) The reaction: 4N₂O + CH₄ $\overset{}{\to}$ CO₂ + 2H₂O + 4N₂ takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N₂O) required to give 150 cm³of steam.
(N = 14, O = 16, C = 12, H = 1)

The reaction : $4 N_{2} O + C H_{4} \to C O_{2} + 2 H_{2} I + 4 N_{2}$ takes place in hte gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide $(N_{2} O)$ required to give $150 c m^{3}$ of steam.

The reaction: takes place in the gaseous state. If all volumes are measured at the same temperature and pressure,

Calculate the volume of dinitrogen oxide (N₂O) required to give 150 cm³ of steam.

[N = 14,O = 16,C=12,H=1]

Q. A mixture of an alkane and

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24

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4N2O+CH4→CO2+2H2O+4N2 If all volumes are measured at the same temperature and pressure, the volume of N2O required to give 150cm3 of steam is :

The correct option is D 300 ccReaction:4N2O+CH4→CO2+2H2O+4N2 4 1 1 2 44 volumes of N2O will give 2 volumes of steam.So, 300 cc of N2O will give 150 cc steam.

$4 N_{2} O + C H_{4} \to C O_{2} + 2 H_{2} O + 4 N_{2}$
If all volumes are measured at the same temperature and pressure, the volume of $N_{2} O$ required to give $150 c m^{3}$ of steam is :

The correct option is D 300 cc
Reaction:
$4 N_{2} O + C H_{4} \to C O_{2} + 2 H_{2} O + 4 N_{2}$
4 1 1 2 4
4 volumes of $N_{2} O$ will give 2 volumes of steam.
So, 300 cc of $N_{2} O$ will give 150 cc steam.