300 mL of a gaseous hydrocarbon when burnt i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by: $C_xH_y + (x + y/4)O_2 	o xCO_2 + (y/2)H_2O$ According to Avogadro's Law,at const

Vedclass · Accepted Answer

$C_8H_{18}$

Vedclass · Answer

(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by: $C_xH_y + (x + y/4)O_2 	o xCO_2 + (y/2)H_2O$ According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to the number of moles. Given: Volume of $C_xH_y = 300 \ mL$ Volume of $CO_2 = 2.4 \ L = 2400 \ mL$ Volume of $H_2O = 2.7 \ L = 2700 \ mL$ From the stoichiometry: $x = 	ext{Volume of } CO_2 / 	ext{Volume of } C_xH_y = 2400 / 300 = 8$ $y/2 = 	ext{Volume of } H_2O / 	ext{Volume of } C_xH_y = 2700 / 300 = 9$ $y = 9 	imes 2 = 18$ Therefore,the molecular formula is $C_8H_{18}$.

$300 \ mL$ of a gaseous hydrocarbon when burnt in excess of $O_2$ gave $2.4 \ L$ of $CO_2$ and $2.7 \ L$ of water vapour under same conditions. The molecular formula of the hydrocarbon is:

Similar Questions

The volume of $0.6 \, M$ $NaOH$ required to neutralize $30 \, cm^3$ of $0.4 \, M$ $HCl$ is ........ $cm^3$.

How many grams of dibasic acid (mol. wt. $200$) should be present in $100 \ mL$ of its aqueous solution to give decinormal strength?

If $0.5 \ g$ of a mixture of two metals $A$ and $B$ with respective equivalent weights $12$ and $9$ displace $560 \ mL$ of $H_2$ at $STP$ from an acid,the composition of the mixture is:

If $5.85 \ g$ of sodium chloride is added to water to make a $500 \ mL$ solution,what is the molarity of the sodium chloride solution (in $M$)? (Molar mass $= 58.5 \ g/mol$)

For preparing one litre $N/10$ solution of $H_2SO_4$,we need $H_2SO_4$.......$g$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module