The question is incomplete, the complete question is;
Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.
Answer:
71.1%
Step-by-step explanation:
The balanced reaction equation must first be written;
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
Let us obtain the number of moles of carbon dioxide corresponding to 28.2 g
Number of moles = mass/ molar mass= 28.2/44.01 g/mol = 0.64 moles of CO2
Next, we obtain the limiting reactant. This is the reactant that yields the least amount of product.
For methane;
Number of moles in 15.1 g= mass/molar mass= 15.1/16gmol-1 = 0.9 moles
From the reaction equation;
1 mole of methane yields 1 mole of carbon dioxide
Hence 0.9 moles of methane yields 0.9 moles of carbon dioxide.
For oxygen
Number of moles of oxygen corresponding to 81.2 g of oxygen= mass/ molar mass= 81.2g/32gmol-1 = 2.5 moles of oxygen
From the reaction equation;
2 moles of oxygen gas yields 1 mole of carbon dioxide
2.5 moles of oxygen gas yields 2.5 × 1 /2 = 1.25 moles of carbon dioxide.
Methane is the limiting reactant.
Theoretical yield of carbon dioxide= 0.9 moles
Actual yield of carbon dioxide= 0.64 moles
% yield = actual yield/ theoretical yield × 100
% yield= 0.64/0.9 ×100
% yield = 71.1%