find the volume of O2 at STP required for the complete combustion of 4g CH4 i)5.6L ii)2.88L iii)22.4L iv)11.2

Open in App

Solution

The explanation for the correct option:
Option(iv) 11.2 L
The combustion reaction of methane is given as: $C H_{4} (g) + 2 O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (l)$
Number of moles of $C H_{4} = \frac{g i v e n m a s s}{m o l a r m a s s} = \frac{4}{16} = 0.25 m o l e s$
From the equation it is clear that one mole of methane reacts with 2 moles of oxygen
Therefore, 0.25 moles of methane will react with $= 0.25 \times 2 = 0.5 m o l e s o f O_{2}$
One mole of an ideal gas occupies 22.4L of volume at STP
Therefore, the volume of oxygen at STP $= 0.5 \times 22.4 = 11.2 L$
Hence, Option(iv) 11.2 L is the volume occupied by oxygen.

Suggest Corrections

142

Similar questions

32

C H_{4}

C H_{4}

16

3 L

C H_{4}

C_{2} H_{6}

9 L

O_{2}

C H_{4}

C_{2} H_{6}

0.25 {d m}^{3}

200 c m^{3}

C_{2} H_{2} + 5 O_{2} \to 4 C O_{2} + 2 H_{2} O

Join BYJU'S Learning Program