Volume of dinitrogen oxide produced at the same time as 8.96 litres of steam is:

22.4 L

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8.96 L

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4.48 L

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11.2 L

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Solution

The correct option is C 4.48 L
$N H_{4} N O_{3} ⟶ N_{2} O + 2 H_{2} O$
1 1 2
1 volume of $N_{2} O$ is equivalent to 2 volumes of steam.

So, 8.96 L steam is equivalent to $\frac{8.96}{2}$ L of $N_{2} O = 4.48$ L of $N_{2} O$ .

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Similar questions

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP).

NH₄NO₃ → N₂O +2H₂O

What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)
NH₄NO₃ $\overset{}{\to}$ N₂O + 2H₂O
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam?
(Relative molecular mass of ammonium nitrate is 80.)
(iii) Determine the percentage of oxygen in the ammonium nitrate. (O = 16)

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP).

NH₄NO₃ → N₂O +2H₂O

What mass of ammonium nitrate should be heated to produce 8.96 litres of steam? (Relative molecular mass of ammonium nitrate is 80.)

Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide? The equation for the reaction is:

3CuO + 2NH₃ → 3Cu + 3H₂O

Volume occupied by 1 mole of gas at STP is 22.4 litres).

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.

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Volume of dinitrogen oxide produced at the same time as 8.96 litres of steam is:

The correct option is C 4.48 LNH4NO3⟶N2O+2H2O 1 1 21 volume of N2O is equivalent to 2 volumes of steam.So, 8.96 L steam is equivalent to 8.962L of N2O=4.48L of N2O.

The correct option is C 4.48 L
$N H_{4} N O_{3} ⟶ N_{2} O + 2 H_{2} O$
1 1 2
1 volume of $N_{2} O$ is equivalent to 2 volumes of steam.

So, 8.96 L steam is equivalent to $\frac{8.96}{2}$ L of $N_{2} O = 4.48$ L of $N_{2} O$ .