Question

The reaction for the combustion of methane is shown above. How many moles of oxygen gas will be consumed in the combustion of 15 moles of methane? Enter your answer using 2 significant figures.
CH4(g) + 2O2(g) => CO2(g) + 2H2O

Answer 1

Answer:

Answer

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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.