Question

30 ml gaseous mixture of methane and ethylene in volume ratio X : Y requires 350 ml air containing 20% of O2 by volume for complete combustion. If ratio of methane and ethylene changed to Y : X. What will be volume of air (in ml) required for complete reaction under similar condition of temperature and pressure.


✔ explanation should be in brief
✯no spam answers
✔answers will be deleted if,
✯irrelevant answer
✯copied
✯spammed
✯additional information required.​

_______________​​​​


​

Answer 1

Solution :-

$\sf{CH_4(g)+2O_2(g) \longrightarrow CO_2(g)+2H_2O(l)}$

$\sf{V_1ml}$        $\sf{2V_1ml}$         $\sf{V_1ml}$            $\sf{0}$

$\sf{C_2H_4(g)+3O_2(g) \longrightarrow 3CO_2(g)+2H_2O(l)}$

$\sf{V_2ml}$        $\sf{2V_2ml}$        $\sf{2V_2ml}$           $\sf{0}$

For the given data,

• $\sf{V_1+V_2=30}$               ... (i)

• $\sf{2V_1+3V_2}$

$\longrightarrow \sf{2V_1+3V_2=350 \times \dfrac{20}{100} }$

$\longrightarrow \sf{2V_1+3V_2=70}$        ...(ii)

Solving equations (i) and (ii) :-

$\sf{V_1=30-V_2}$

$\longrightarrow \sf{ 2(30-V_2)+3V_2=70 }$

$\longrightarrow \sf{ 60-2V_2+3V_2=70}$

$\longrightarrow \sf{ V_2=10}$

$\longrightarrow \sf{ V_1=30-10}$

$\longrightarrow \sf{ V_1=20}$

$\sf{ \therefore V_1=10\:\:and\:\:V_2=20}$

→ Volume of O₂ required = 2V₁ + 3V₂ = 80 ml

Volume of air required = 80 × 100/20

→ 80 × 5 = 400 ml

◘ Therefore, the answer is 400 ml.

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\dashrightarrow V_1+V_2=30\:---\boxed{1}$

$\dashrightarrow 2V_1+3V_2=350 \times \dfrac{20}{100}$

$\dashrightarrow 2V_1+3V_2=70\:---\boxed{2}$

$\dashrightarrow CH_4(g)+2O_2(g) \dashrightarrow CO_2(g)+2H_2O(l) \\ v_1ml,2v_1ml, v_1ml :0$

$\dashrightarrow C_2H_4(g)+3O_2(g) \dashrightarrow 3CO_2(g)+2H_2O(l)\\ v_2ml,2v_2ml,2v_2ml:0$

$\therefore solving\:1\:and\:2.we\:get,$

$\implies V_1=30-V_2$

$\implies 2(30-V_2)+3V_2=70$

$\implies 60-2V_2+3V_2=70$

$\implies V_2= 70-10$

$\implies V_2=10$

$\large{\boxed{\bf{ \star\:\: V_2=10\:\: \star}}}$

$\dashrightarrow V_2=10 \\ \dashrightarrow V_1=30-V_2\\ \dashrightarrow 30-10\\ \implies V_1=20$

$\large{\boxed{\bf{ \star\:\: V_1=20\:\: \star}}}$

$\therefore Volume\:of\:O_2= 2v_1+3v_2= 2(20)+3(10)= 40+30=70ml \\dashrightarrow volume\:of\:air\:= 70\times \dfrac{100}{20} \\ \implies 70\times \dfrac{\cancel{100}}{\cancel{20}} \\ \implies 70\times 5 \\ \implies 350 ml$

____________