Methane gas combusts according to the following chemical equation:
CH4 (8) + 202(g) → CO2(g) + 2H2O(l)
29.2 L of methane gas is combusted with 63.3 L of
oxygen gas at STP. What volume of carbon dioxide
is produced in the reaction?
___L CO2
Your answer should be rounded to three significant figures. Do
not include units in your answer.

The volume of carbon dioxide is produced in the reaction is 29.12 L

We know that

1 mol gas at STP = 22.4 L

So,

The mol methane is

Now

The  mol oxygen is

Now the volume is

= 29.12 L

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The volume of Carbon dioxide : 29.12 L

Given

CH4 (g) + 202(g) → CO2(g) + 2H2O(g)

Oxygen volume = 63.3 L

Methane volume = 29.2 L

The volume of Carbon dioxide

Solution

1 mol gas at STP = 22.4 L

mol methane :

mol oxygen :

limiting reactant : CH₄(smaller ratio:mol/reaction coefficient)

So mol CO₂ based on CH₄

From equation, CO₂ : CH₄ = 1 : 1, so mol CO₂ = 1.3

The volume =

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explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.