Question

The combustion of propane may be described by the chemical equation C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g)O2(g) are needed to completely burn 83.3 g C3H8(g)?

Answer 1

302.22 g O2(g) are required

Step-by-step explanation:

• C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

∴ Mw C3H8 = 44.1 g/mol

∴ Mw O2 = 32 g/mol

⇒ g O2 = (83.3g C3H8)×(mol C3H8/44.1g C3H8)×(5mol O2/mol C3H8)×(32g O2/mol O2)

⇒ g O2 = 302.22 g O2