Consider the reaction for the combustion of methanol (CH3OH): 2CH3OH+3O2⟶2CO2+4H2O What is the mass of oxygen...

Question

Question

Consider the reaction for the combustion of methanol (CH3OH):

2CH3OH+3O2⟶2CO2+4H2O

What is the mass of oxygen (O2) that is required to produce 579g of carbon dioxide (CO2)?

Answer 1

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Answer #1

From the equation given, we can say that, 2 moles of methanol reacts with 3 moles of O₂ to produce 2 moles of CO₂ and 4 moles of H₂O.

Now, molar mass of CO₂ is 44 g.mol^-1.

Thus, 579 g CO₂ = (579 / 44) mol CO₂ = 13.1591 mol

Since from 3 moles of O₂ , 2 moles of CO₂ is produced, thus moles of O₂ reacted in this case

= (13.1591 x 3 / 2) mol = 19.7386 mol.

Molar mass of O₂ is 32 g.mol^-1

Thus, mass of O₂ needed is (19.7386 x 32) g = 631.64 g.

Answer 2

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