Question

In a saturated acyclic compound the mass ratio C:H His 4: 1 and C:O 3: 4. Find the no. of moles of O2 required to react with 2 moles compound to give CO2 and water.

Answer 1

GIVEN:-

• In a saturated acyclic compound the mass ratio C:H is 4: 1 and C:O is 3: 4.

TO FIND :-

• The no. of moles of O2 required to react with 2 moles compound to give CO2 and water.

SOLUTION:-

Given that , C:H = 4:1

C:O = 3:4

》 Molar ratio of C:H = 12 :1 .

$\implies$ Mass ratio of C & H / Molar mass ratio = 4 : 1 / 12 : 1

$\implies$ 4 : 12 = 1 : 3.

$\rightarrow$ CH3

》 C : O = 12 : 16

$\implies$3:4/12 : 16 = 3:4/3:4 = 1.

$\rightarrow$ CO

Hence , 1 Mole of O2 is required.

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Answer 2

★ QUESTION :

In a saturated acyclic compound the mass ratio, C:H His 4: 1 and C:O 3: 4. Find the number of moles of O2 required to react with 2 moles compound to give CO2 and water.

GIVEN :

• The mass ratio C : H = 4:1 and C:O 3:4

TO FIND :

• Number of moles of O2 (oxygen) required to react with two moles of compound to give CO2 (carbon dioxide) and water (H2O) = ?

STEP -BY-STEP EXPLAINATION :

Here as the provided question, is saying :

• C : H = 4 : 1 and C:O = 3:4

So, now we will have to find here ratio (molar) C :H = ?

Hence, we'll apply here are the formula for finding molar ratio :

ㅤ⟹ C : H = value of C : H in provided question × value of C : O (Given)

Substituting here the values, as per the given formula :

➠ C : H = C : H × C : O

➠ C : H = 4 : 1 × 3 : 4

➠ C : H = 12 : 1

Now, we got the value of molar ratio (C:H) = 12 : 1

$➠ \frac{ᴍᴏʟᴀʀ \: ʀᴀᴛɪᴏ \: ᴏғ \: ᴄ \: ᴀɴᴅ \: ʜ }{ᴍᴏʟᴀʀ \: ᴍᴀss \: (ʀᴀᴛɪᴏ) \: }$

Substituting here the values, as per the given formula :

$➠ \frac{4:1}{12:1}$

After reducing 4 :1 / 12 :1 we will get here 3: 1

$➠ (ᴄᴏɴᴠᴇʀᴛɪɴɢ \: ʜᴇʀᴇ \: ᴛʜᴇ \:ᴠᴀʟᴜᴇ \: ᴀs \: 4: 12)$

After reducing, 4:12, (with the table of 4)

We, got here the value as 1 : 3

→ CH3

Now, finding the value, so that we can gain the the required the mole of Oxygen that is to be reacted with two moles compound 2 minut carbon dioxide and water :

ㅤㅤ⟹ C : O = 12 : 6 (gain value)

$⟹ \frac{3: 4}{12: 16} \:$

After reducing here, we got :

$⟹ \frac{3:4}{3:4}$

When 3:4 and 3:4 is reduced, we got here the reduced value as 1

→ CO

Therefore, 1 mole of Oxygen is required to react with two moles compound to give carbon dioxide and water.