Answer:
174.4 g of oxygen gas required
Step-by-step explanation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
In a chemical reaction, the products and reactants always remain in stoichiometric ratios, and therefore:
C₃H₈ : 5O₂ : 3CO₂ : 4H₂O = 1 : 5 : 3 : 4
Therefore, we need 5 moles of O₂ gas to react with 1 mole of C₃H₈.
To calculate number of moles (n), we can divide mass in grams (m) by molar mass (g/mol). Molar mass (MM) is found on a standard IUPAC Periodic Table.
n(C₃H₈) = m/MM = 48.05/(12.01×3+1.008×8) = 1.089717 mol
Now, using our stoichiometric ratios, 1 : 5, n(O₂) = 5×n(C₃H₈).
∴n(O₂) = 5×1.089717 = 5.448587 mol
To find the mas required, we can rearrange the above formula from before that we used to calculate the number of moles of C₃H₈. i.e, multiply moles by molar mass to get mass present.
m(O₂) = n×MM = 5.448587×(16.00×2) = 174.4 g