How many grams of methane should be burned in an excess of oxygen at stp to obtain 5.6 l of carbon dioxide?

The balanced reaction that describes the reaction between methane and oxygen to produce carbon dioxide and water is expressed Ch4 +2O2 = CO2 +2H2O. We use PV=nRT to determine the moles of carbon dioxide equivalent to 5.6 L of it. For every mole of ideal gas, there is an equivalent volume of 22.4 L. Hence there are 0.25 moles of CO2. The mass of methane used is 4 grams methane.

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BarrettArcher BarrettArcher · Answer 2 · 2019-01-30T06:53:13+01:00

Answer : The amount of methane burned is, 4 grams.

Solution : Given,

Volume of carbon dioxide = 5.6 L

Molar mass of methane = 16 g/mole

First we have to calculate the moles of methane.

At STP, 1 mole of gas contains 22.4 L volume of gas.

As, 22.4 L volume of carbon dioxide obtained from 1 mole of carbon dioxide gas

As, 5.6 L volume of carbon dioxide obtained from [tex]\frac{5.6}{22.4}=0.25[/tex] moles of carbon dioxide gas

Now we have to calculate then moles of methane.

The balanced chemical reaction will be,

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

From the reaction we conclude that

1 mole of carbon dioxide produces from 1 mole of methane

0.25 mole of carbon dioxide produces from 0.25 mole of methane

The moles of methane gas = 0.25 moles

Now we have to calculate the mass of methane.

[tex]\text{Mass of }CH_4=\text{Moles of }CH_4\times \text{Molar mass of }CH_4[/tex]

[tex]\text{Mass of }CH_4=(0.25moles)\times (16g/mole)=4g[/tex]

Therefore, the amount of methane burned is, 4 grams.