10 mL of a gaseous hydrocarbon on combustion | 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:$C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow xCO_2 + \frac{y}{2} H_2O$According to A

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by:$C_xH_y + (x + \frac{y}{4}) O_2 ightarrow xCO_2 + \frac{y}{2} H_2O$According to Avogadro's Law,the volume of gases is proportional to the number of moles.Given: $10 \ mL$ of $C_xH_y$ produces $40 \ mL$ of $CO_2$ and $50 \ mL$ of $H_2O$.From the stoichiometry,$1 \ mol$ of $C_xH_y$ produces $x \ mol$ of $CO_2$ and $\frac{y}{2} \ mol$ of $H_2O$.Thus,$10x = 40 \implies x = 4$.And $10 	imes (\frac{y}{2}) = 50 \implies 5y = 50 \implies y = 10$.The hydrocarbon is $C_4H_{10}$.Total number of atoms $= x + y = 4 + 10 = 14$.

$10 \ mL$ of a gaseous hydrocarbon on combustion gives $40 \ mL$ of $CO_{2(g)}$ and $50 \ mL$ of water vapour. The total number of carbon and hydrogen atoms in the hydrocarbon is:

Similar Questions

If the density of iron is $7.86 \, g \, cm^{-3}$ and its atomic mass is $55.85 \, u$,then find the volume occupied by $1 \, mol$ of iron atoms in $cm^{3}$.

The normality of $1 \ M$ solution of $H_3PO_4$ will be.....$N$

How many moles of ferric alum $(NH_4)_2SO_4 \cdot Fe_2(SO_4)_3 \cdot 24H_2O$ can be made from a sample of $Fe$ containing $0.0056 \ g$ of it?

What is the volume of $O_2$ gas required at $STP$ for the combustion of $4 \ g$ of methane gas (in $L$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module