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

a graduated cylinder.

explanation:

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.