40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@) C and 1 atm . A spark was produced so that the formation of water was complete . The remaining pure gas had a volume of 10 mL of 18^(@) C & 1 atm . If the remaining gas was H_(2) what was initial mole % of H_(2) in mixture ?

40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@

1.	40 mL of mixture of H_(2) & O_(2) was placed in a gas burette at 18^(@) C and 1 atm . A spark was produced so that the formation of water was complete . The remaining pure gas had a volume of 10 mL of 18^(@) C & 1 atm . If the remaining gas was H_(2) what was initial mole % of H_(2) in mixture ?
Answer» <html><body><p>0.75<br/>0.25<br/>0.6<br/>0.45</p>Solution :Let the volume of `O_(2)` is the mixture be x mL <br/> `H_(2) + 1//2 O_(2) to H_(2) O (l) ` <br/> <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> vol. in ml `"" (40 - x)"" x` <br/> <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> vol. in ml (40- <a href="https://interviewquestions.tuteehub.com/tag/3x-310805" style="font-weight:bold;" target="_blank" title="Click to know more about 3X">3X</a>) - 0. <br/> Final volume of `H_(2) = 10 mL ""` (<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a>) <br/> `therefore 40 - 3x = 10 , x = 10 ` mL <br/> `therefore` initial volume of `H_(2) = 30` mL , Initial mole % of `H_(2) = (30)/(40) xx 100 = 75%`</body></html>