A definite amout of gaseous hydrocarbon was burnt with just sufficient of O_(2).The volume of all reactantswas 600ml,after the explosion the volume of the products [CO_(2)(g) and H_(2)O(g) was found to be 700ml under the similar conditions the molecular formula of the compound is .

A definite amout of gaseous hydrocarbon was burnt with just sufficient

1.	A definite amout of gaseous hydrocarbon was burnt with just sufficient of O_(2).The volume of all reactantswas 600ml,after the explosion the volume of the products [CO_(2)(g) and H_(2)O(g) was found to be 700ml under the similar conditions the molecular formula of the compound is .
Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)H_(8)`<br/>`C_(3)H_(6)`<br/>`C_(3)H_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`<br/>`C_(4)H_(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)`</p>Solution :Mass `%` of `O=((<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>/(98)xx4xx16+(<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>)/(80)xx3xx16)xx100)/(100) = 63.18%`</body></html>