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) ?

To completely burn 83.3 grams of propane, there is 302.24 grams O2 needed.

Explanation:

Step 1: Data given

Mass of C3H8 = 83.3 grams

Molar mass of C3H8 = 44.1 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

C3H8 (g) + 5O2 (g) ⟶3CO2 (g) + 4H2O (g)

Step 3: Calculate moles of C3H8

Moles C3H8 = mass of C3H8 / Molar mass C3H8

Moles C3H8 = 83.3 grams / 44.1 g/mol

Moles C3H8 = 1.889 moles

Step 4: Calculate moles O2

For 1 mole C3H8 consumed, we need 5 moles of O2

For 1.889 moles of C3H8, we need 5 * 1.889 = 9.445 moles of O2

Step 5: Calculate mass of O2

Mass O2 = moles O2 * Molar mass O2

Mass O2 = 9.445 moles * 32 g/mol

Mass O2 = 302.24 grams O2

To completely burn 83.3 grams of propane, there is 302.24 grams O2 needed.

302.22 g O2 (g) are required

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