MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

4.3 g of satu...

Question

4.3 g of saturated hydrocarbon $(C_{n} H_{(2 n + 2)})$ is burnt completely and all $C O_{2}$ were absorbed in aqueous $N a O H$ solution. A 300 mL of a 2 M $N a O H$ solution was required for complete absorption of $C O_{2}$ . Hence find the value of n.

Open in App

Solution

We know,
$2 N a O H + C O_{2} \to N a_{2} C O_{3} + H_{2} O$
So 2 mol $N a O H$ absorbs 1 mol of $C O_{2}$
Here we have,
300 mL of a 2 M $N a O H$ = 600 mmol of $N a O H$
So, 600 mmol of $N a O H$ will absorb 300 mmol of $C O_{2}$

So, we got 300 mmol of $C O_{2}$ by burning 4.3 g of $C_{n} H_{(2 n + 2)}$
So applying POAC on carbon we get,
$\frac{4.3}{14 n + 2} = \frac{0.3}{n}$
So, $n = 6$

Suggest Corrections

thumbs-up

3

Similar questions

One mole of a mixture of CO and CO₂ requires exactly 20 g of NaOH in solution for complete conversion of all the CO₂ into Na₂CO₃. How much NaOH would it require for conversion into Na₂CO₃, if the mixture (one mole) is completely oxidized to CO₂?

Q. One mole of a mixture of

C O

C O_{2}

requires exactly

20

N a O H

in solution for complete conversion of all the

C O_{2}

N a_{2} C O_{3}

. If the mixture (one mole) is completely oxidised to

C O_{2}

, then weight of

N a O H

Q. 0.8 mole of a mixture of

C O

C O_{2}

requires exactly 40 gram of

N a O H

in solution for complete conversion of all the

C O_{2}

N a_{2} C O_{3}

. How many moles more of

N a O H

would it require for conversion into

N a_{2} C O_{3}

, if the mixture (0.8 mole) is completely oxidised to

C O_{2}

:

One mole of a mixture of CO and CO₂ requires exactly 20 g of NaOH in solution for complete conversion of all the CO₂ into Na₂CO₃. How much NaOH would it require for conversion into Na₂CO₃, if the mixture (one mole) is completely oxidized to CO₂?

(H_{2} N C O N H_{2})

is manufactured by passing

C O_{2} (g)

through ammonia solution followed by crystallization.

C O_{2}

for the above reaction is prepared by combustion of hydrocarbon. If combustion of

236 k g

of a saturated hydrocarbon

(C_{n} H_{2 n + 2})

produces as much

C O_{2}

as required for production of

999.6 k g

urea. Then, the molecular formula of the hydrocarbon is :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Stoichiometric Calculations

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Stoichiometric Calculations

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon