Question

CH2, (g) +202, (g) → CO2.(g) + 2H2O(l).
According to given equation if 16 g methane is allowed to react with excess of oxygen the volume of CO2
released at STP will be-
(1) 22.4L
(2) 11.2 L
(3) 44.8L
(4) 5.6 L​

Answer 1

Answer:

CH

4

(g)+2O

2

(g)→CO

2

(g)+2H

2

O(g)

Since 1 mole methane = 12+4×1=16 g and 2 mole oxygen =2×32=64 g, 1 mole CO

2

=12+32=44 g and 2 mole water=2×18=36 g

From the balanced reaction: One mole(or 1 molecule or 16 g) of CH

4

reacts with 2 moles (or 2 molecules or 64 g) of oxygen to give one mole (or 1 molecule or 44 g) of CO

2

and 2 mole (or 2 molecules or 36 g) of water.

For the gaseous system at STP:

22.4 L methane reacts with 44.8 L of oxygen to give 22.4 L of CO

2

and 44.8 L of water

Explanation:

please mark me as brainlist

Answer 2

Answer:

ok is this maths I like this