Combustion of 10 mL of a gaseous hydrocarbon | vedclass

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

Question

(D) According to Avogadro's Law, the volume ratio of gases is equal to the mole ratio under the same conditions of temperature and pressure.Let the hy

Vedclass · Accepted Answer

$C_4H_{10}$

Vedclass · Answer

(D) According to Avogadro's Law, the volume ratio of gases is equal to the mole ratio under the same conditions of temperature and pressure.Let the hydrocarbon be $C_xH_y$.The combustion reaction is: $C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O$.Given volume ratio: $V(C_xH_y) : V(CO_2) : V(H_2O) = 10 : 40 : 50 = 1 : 4 : 5$.From the stoichiometry, $x = 4$ (moles of $CO_2$ per mole of hydrocarbon).Also, $y/2 = 5$, which implies $y = 10$.Therefore, the molecular formula is $C_4H_{10}$.

Combustion of $10 \ mL$ of a gaseous hydrocarbon gives $40 \ mL$ of $CO_2$ and $50 \ mL$ of water vapour under the same conditions. The molecular formula of the hydrocarbon is

Similar Questions

The minimum amount of $O_2\,(g)$ consumed per gram of reactant is for the reaction: (Given atomic mass: $Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1$)

When $100 \, ml$ of $0.1 \, M$ $H_2SO_4$ reacts with $Na_2CO_3$, how many liters of $CO_2$ will be produced at $STP$ (in $L$)?

How much water is needed to dilute $10 \, mL$ of $10 \, N$ hydrochloric acid to make it exactly decinormal $(0.1 \, N)$? .......... $mL$

What does the equation $2Al(s) + \frac{3}{2}O_2(g) \to Al_2O_3(s)$ represent?

The volume of water to be added to $100 \, cm^3$ of $0.5 \, N \, H_2SO_4$ to get decinormal concentration is ........ $cm^3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module