Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . what is the theoretical yield of carbon dioxide formed from the reaction of of methane and of oxygen gas?

The theoretical yield of carbon dioxide formed from the reaction of methane and oxygen gas is 1 mol.

Explanation:

The equation below shows the reaction of combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

When no reactants are wasted and the reaction is complete, 1 mol of gaseous CO₂ will be produced. This means that the theoretical yield of carbon dioxide is 1 mol.

Completed question:

Gaseous methane  ( C H 4 )  reacts with gaseous oxygen gas  ( O 2 )  to produce gaseous carbon dioxide  ( C O 2 )  and gaseous water  ( H 2 O ) . What is the theoretical yield of carbon dioxide formed from the reaction of   0.16  g of methane and  0.83  g  of oxygen gas?

0.44 g of CO₂

Explanation:

The reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

So, by the equation, we can notice that 1 mol of methane (CH₄) reacts with 2 moles of oxygen (O₂) to form 1 mole of carbon dioxide (CO₂) and 2 moles of water (H₂O).

The yield of a reaction is how much of the product is formed when a determined quantity of the reagents reacts. In the real-life, some complications can happen, such as parallel reactions, problems in pressure, temperature, or even in the reactor, and so the real yield will be less than the one predicted by the reaction (theoretical yield).

First, let's identify which one of the reagents is the limiting, which is, the one that is totally consumed, and so, will be the one that will conduct the yield.

The stoichiometry between the two reagents is 1 mol of CH₄ to 2 moles of O₂. The molar mass of CH₄ is 16 g/mol, and of O₂ is 32 g/mol, thus multiplying the number of moles by the molar mass, the ratio between them is:

16 g of CH₄/64 g of O₂

According to the Proust Law, this ratio must be constant. So, if there is 0.16 g of methane, the mass of oxygen must be:

16/32 = 0.16/m

16m = 0.16*32

16m = 5.12

m = 0.32 g of O₂

Because there is 0.83 g of O₂ it is in excess and CH₄ is the limiting reagent.

The stoichiometry between CH₄ and CO₂ is 1 mol of CH₄ to 1 mol of CO₂, the molar mass of CO₂ is 44 g/mol, so the mass rate is:

16 g of CH₄/44 g of CO₂

And the mass formed of CO₂:

16/44 = 0.16/x

16x = 44*0.16

16x = 7.04

x = 0.44 g of CO₂