In the present case, V∝n(∵ all the volumes are measured under identical conditions of temperature and pressure) Hence, the reaction Stoichiometry can be solved using volumes as:Cx​Hy​( g)+(x+4y​)O2​( g)⟶xCO2​( g)+2y​H2​O(l)volume of CO2​(g)+O2​( g) (remaining unreacted) =25⇒ Volume of CO2​( g) produced=10 mL(15 mLOO2​ remaining )∵1 mLCx​Hy​ produces x mL of CO2​∴5 mLCx​Hy​ will produce 5x mL of CO2​=10 mL⇒x=2Also, 1 mLCx​Hy​ combines with (x+4y​)mL of O2​5 mLCx​Hy​ will combine with 5(x+4y​)mL of O2​⇒5(x+4y​)=15(15 mL of O2​ out of 30 mL) (remaining unreacted)⇒y=4, hence hydrocarbon is C2​H4​.