Question

2 moles of a mixture of acetylene and ethane produced on combustion, 5500 KJ of heat energy. The
enthalpies of combustion of acetylene and ethane are 2000 and 3000 kJ mol-" respectively. The ratio of
acetylene to ethane in the mixture is
1) 1:3
2) 1:1
3) 3:1
4) 1:2
​

Answer 1

Answer:

1:3

Explanation:

Let number of moles of acetylene be and no.of moles of ethane be 1-x.

If one mole acetylene produces 2000KJ on combustion, x moles produce 2000x KJ;

Similarly, if one mole ethane produce 3000KJ then 1-x moles produce 3000(1-x) KJ energy.

Given that 2 moles of the mixture produce 5500KJ then one mole produce 2750KJ.

Now, 2000x + 3000(1-x) = 2750

On soving x=1/4 and 1-x = 3/4.

Taking the ratio = 1:3.

Ps:-not so sure about it..!!!"please check with some other resources also"!!!

Answer 2

Answer:

The ratio of acetylene to ethane in the mixture is 1 : 3.

Therefore, option (1) is correct.

Explanation:

Consider that 'x' is the number of moles of acetylene and '1-x' is the number of moles of ethene.

Given, for one mole enthalpies of combustion of acetylene = 2000KJ

For x moles  of acetylene the enthalpies of combustion = $2000x$ KJ

Similarly, for one mole enthalpies of combustion of ethane = 3000 KJ

For (1-x) mole of acetylene, enthalpies of combustion = $3000(1-x)$ KJ

Given that 2 moles of the mixture produce heat energy = 5500KJ

Then one mole will produce heat energy = 2750KJ.

For the mixture of acetylene and ethane

$Now,\; 2000x + 3000(1-x) = 2750$

$2000x + 3000-3000x = 2750$

$-1000x = -250$

$x =1/4$

Then, $1-x = 1-\frac{1}{4}=\frac{3}{4}$

The ratio of acetylene to ethane in the mixture = $\frac{1}{4} :\frac{3}{4}$

Simplify, the ratio is 1 : 3

Therefore, the ratio of acetylene to ethane in the mixture is equal to 1 : 3.

To know more about "Combustion of ethane"

https://brainly.in/question/38429443

To know more about "gas has the highest heat of combustion"

https://brainly.in/question/3504925