Respuesta :
Answer:
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.
Answer:
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
