Answer:
Theoretical yield = 3.51 g
Explanation:
The formula for the calculation of moles is shown below:
 
For  :-
  :- 
Mass of  = 1.28 g
  = 1.28 g
Molar mass of  = 16.04 g/mol
  = 16.04 g/mol
The formula for the calculation of moles is shown below:
 
Thus,
 
 
For  :-
  :-
Mass of  = 10.1 g
  = 10.1 g
Molar mass of  = 31.998 g/mol
  = 31.998 g/mol
The formula for the calculation of moles is shown below:
 
Thus,
 
 
According to the given reaction:
 
1 mole of methane gas reacts with 2 moles of oxygen gas
0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas
Moles of oxygen gas = 0.1596 moles
Available moles of oxygen gas = 0.3156 moles
<u>Limiting reagent is the one which is present in small amount. Thus,  is limiting reagent.
</u>
 is limiting reagent.
</u>
The formation of the product is governed by the limiting reagent. So,
1 mole of methane gas on reaction produces 1 mole of carbon dioxide.
0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.
Mole of carbon dioxide = 0.0798 mole
Molar mass of carbon dioxide = 44.01 g/mol
The formula for the calculation of moles is shown below:
 
Thus,
 
Mass of  = 3.51 g
 = 3.51 g
<u>
Theoretical yield = 3.51 g</u>