contestada

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.

Respuesta :

ardni313

The volume of Carbon dioxide : 29.12 L

Further explanation

Given

Reaction

CH4 (g) + 202(g) → CO2(g) + 2H2O(g)

Oxygen volume = 63.3 L

Methane volume = 29.2 L

Required

The volume of Carbon dioxide

Solution

1 mol gas at STP = 22.4 L

mol methane :

[tex]\tt \dfrac{29.2}{22.4}=1.3[/tex]

mol oxygen :

[tex]\tt \dfrac{63.3}{22.4}=2.83[/tex]

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 =

[tex]\tt 1.3\times 22.4=29.12~L[/tex]

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

  • The calculation is as follows:

We know that

1 mol gas at STP = 22.4 L

So,

The mol methane is

[tex]= 29.2 \div 22.4\\\\= 1.3[/tex]

Now

The  mol oxygen is

[tex]= 63.3 \div 22.4\\\\= 2.83[/tex]

Now the volume is

[tex]= 22.4 \times 1.3[/tex]

= 29.12 L

Learn more: https://brainly.com/question/5763151?referrer=searchResults

Otras preguntas