Question

calculate the volume of O2 at NTP required burning 30 gram of CH4​

Answer 1

Answer:

Explanation:

CH4 + 2 O2 → CO2 + 2H2O

1 mole of CH4 needs 2 moles of molecular O2.

You have 30 g of CH4 or 30 g/(16 g/mole) = 1.875 mole of CH4. Since mole ratio between CH4 and O2 = 1 : 2, then, 1.875 mole of CH4 stoichiometrically reacts with 2/1 x 1.875 mole of O2 = 3.75 mole of O2.

the volume of O2 required to completely combust CH4 at STP condition = 3.75 mole x 22.4 L/mole = 84 L.

Answer 2

Answer:

84 L

Explanation:

Reaction between C$H_4$ and $O_2$ is given as -

             C$H_4$  +  2$O_2$ ⇒  C

Moles of  $O_2$  required to react with 1 mole of C$H_4$ = 2

Given mass of C$H_4$ = 30g

so, no. of moles of C$H_4$ = 30/16 = 1.875

So. no. of moles of  $O_2$ required = 2 × 1.875 = 3.75 moles

Now Molecular Volume of  $O_2$ at NTP = 22.4 L

Therefore, volume of  $O_2$ required = 22.4 × 3.75 L = 84 L

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