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how many litres of detonating gas will be produced at ntp in the decomposition of 0.1 mole of waterby an electric current?

how many litres of detonating gas will be produced at ntp in the decomposition of 0.1 mole of waterby an electric current?

Grade:12

4 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

Use the T and P conditions of NTP...

ankitesh gupta
63 Points
11 years ago

AT STP AND NTP

n(MOLES)=V(LITRES)/22.4

ACCORDING TO GIVEN INFORMATION n=0.1 V=?

0.1 X 22.4=V

V=2.24LITRES........................ANSWER

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Harshad Dighe
21 Points
7 years ago
2H2O------->2H2 +O2
Detonating  gas means H2 and O2 
therefore total no of moles= 3
2mol-------> 3 mole
0.1mol-------> X mole
X=0.5
no. of mol= volume/22.4........(at ntp)
therefore volume=0.5 × 22.4= 3.36 litre
Kan Haiya Gupta
26 Points
7 years ago
Decomposition of water by electric current can be represented by the following reaction:2H2O (l) → 2H2 (g) + O2 (g)2 moles 2 moles 1 moleNow, the mixture of H2 (g) and O2 (g) is known as the detonating gas. So, total number of moles of detonating gas produced, from decomposition of 2 mole of water = (2 mole + 1 mole) = 3 moles.Now, therefore 0.1 mole of water will produce (0.1 x 3) / 2 = 0.15 moles of detonating gas.1 mole of detonating gas at NTP = 22.4 L of the gasTherefore, 0.15 mole of detonating gas contains (22.4 x 0.15) = 3.36 L of detonating gas at NTP.Hope, this information is helpful.

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