What mass of o2 is required to produce 14.5 g of co2 if the reaction has a 65.0% yield? ch4(g) + 2o2(g) \longrightarrow ⟶ co2(g) + 2h2o(g)?

the balanced equation for the combustion of methane is as follows

CH₄ + 2O₂ --> CO₂ + 2H₂O

stoichiometry of O₂ to CO₂ is 2:1

the percentage yield of CO₂ is 65.0 %

percentage yield = (actual yield / theoretical yield) x 100 %

substituting these values to find the theoretical yield

65 % = (14.5 g / theoretical yield) x 100 %

theoretical yield = 14.5 / 0.65 = 22.3 g

the theoretical yield of CO₂ is 22.3 g

therefore number of moles of CO₂ that was expected to be formed - 22.3 g / 44 g/mol = 0.507 mol

according to molar ratio O₂ to CO₂ ratio is 2:1

therefore number of O₂ moles required to react - 0.507 x2 = 1.01 mol

mass of O₂ required - 1.01 mol x 32 g/mol = 32.32 g

therefore mass of O₂ needed is 32.32 g

43.452

step-by-step explanation:

284 times .153 = 43.452

it is stable as it is in group 0 which have full outer shells of electrons