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

Dear student,

Use the T and P conditions of NTP...

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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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.