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

(D) The combustion reaction for a hydrocarbon $C_xH_y$ is: $C_xH_y(g) + (x + \frac{y}{4}) O_2(g) \rightarrow x CO_2(g) + \frac{y}{2} H_2O(g)$ Given vo

Vedclass · Accepted Answer

$C_4H_{10}$

Vedclass · Answer

(D) The combustion reaction for a hydrocarbon $C_xH_y$ is: $C_xH_y(g) + (x + \frac{y}{4}) O_2(g) \rightarrow x CO_2(g) + \frac{y}{2} H_2O(g)$ Given volumes: $10 \ mL$ of hydrocarbon produces $40 \ mL$ of $CO_2$ and $50 \ mL$ of $H_2O$. According to Avogadro's Law,volume is proportional to the number of moles at constant temperature and pressure. Ratio of volumes: $1 : x : \frac{y}{2} = 10 : 40 : 50$ Dividing by $10$: $1 : 4 : 5$ Comparing the coefficients: $x = 4$ $\frac{y}{2} = 5 \Rightarrow y = 10$ Thus,the hydrocarbon is $C_4H_{10}$.

$10 \ mL$ of a gaseous hydrocarbon on combustion gives $40 \ mL$ of $CO_{2(g)}$ and $50 \ mL$ of $H_2O_{(g)}$. The hydrocarbon is $-$

Similar Questions

Calculate the volume of $99 \text{ g}$ of $CO_2$ at $STP$. (in $\text{ L}$)

$0.5 \ L$ of gaseous hydrocarbon when burnt in excess of $O_2$ gave $3.5 \ L$ of $CO_2$ and $4 \ L$ of water vapours under same conditions. The molecular formula of the hydrocarbon will be:

$X \ g$ of nitrobenzene on nitration gave $4.2 \ g$ of $m-$dinitrobenzene. $X = ............ \ g$. $(nearest \ integer)$ [Given: molar mass (in $g \ mol^{-1}$) $C: 12, H: 1, O: 16, N: 14$]

$A$ pre-weighed vessel is filled with $O_2$ and weighed at $STP$. It is then evacuated and filled with $SO_2$ gas at the same temperature and pressure,and weighed again. The weight of $O_2$ will be:

What is the weight of $50\%$ pure $HCl$ required to react completely with $100 \ g$ of $100\%$ pure limestone $(CaCO_3)$ (in $g$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module