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if there are 20g of methane, 64g of Oxygen gas and 44g of carbon dioxide in this combustion reaction. How many grams of water were produce?

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A balanced equation for combustion of methane is given below:

CH

4

(

g

)+2O

2

(

g

)→CO

2

(

g

)+2H

2

O(

g

)

Which of the following statements is not correct on the basis of the above chemical equation?

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Solution

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Correct option is

C

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

2

and 22.4 L of water

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.

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