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1 C3H8 + 5 O2 → 3 CO2 + 4 H2O

What mass of O2 is needed to react with 48.05 g C3H8?

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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

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