Is 400 g of oxygen gas enough to burn 20 moles of methane completely-A- YesB- All the aboveC- NoD- Insufficient data

The correct option is C No The combustion of methane is CH_4+2O_2→ CO_2+2H_2O i.e., One mole of methane requires two moles of oxygen gas. i.e., 1 mole = (2 × ...

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Byju's Answer

Standard XII

Chemistry

Limiting Reagent or Reactant

Is 400 g of o...

Question

Is 400g of oxygen gas enough to burn 20 moles of methane completely?

Yes

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Insufficient data

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All the above

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Solution

The correct option is B
No

The combustion of methane is

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

i.e., One mole of methane requires two moles of oxygen gas.

i.e., $1 m o l e$ = $(2 \times 32) g$ of $O_{2}$

$20 m o l e s$ of $C H_{4}$ = $20 \times (2 \times 32) g$ of $O_{2}$

$20 m o l e s$ of $C H_{4}$ = $1280 g$ of $O_{2}$

i.e., $20 m o l e s$ of $C H_{4}$ will require $1280 g$ of $O_{2}$ to burn completely

Now, since there is only $400 g$ of $O_{2}$ gas, some $C H_{4}$ can-not be burnt and will be left over after the reaction.

So, in this reaction, $O_{2}$ is the limiting reagent as it gets used up completely.

So, in this reaction, $C H_{4}$ is the excessive reagent.

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