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kakasveta [241]
3 years ago
10

What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?

Chemistry
1 answer:
Marta_Voda [28]3 years ago
6 0

Answer : The excess reactant in the combustion of methane in opem atmosphere is O_{2} molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

First we have to calculate the moles of methane.

\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}} = \frac{23g}{16.04g/mole} = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give \frac{2moles\times 1.434moles}{1moles} moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is O_{2} molecule.


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