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Answer:
4.06 mol H₂O
Explanation:
- 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
First we <em>convert the given masses of reactants into moles</em>, using <em>their respective molar masses</em>:
- 250 g O₂ ÷ 32 g/mol = 7.81 mol O₂
- 50 g C₆H₁₄ ÷ 86 g/mol = 0.58 mol C₆H₁₄
Now we <u>calculate how many O₂ moles would react completely with 0.58 C₆H₁₄ moles</u>, using the <em>stoichiometric coefficients of the reaction</em>:
- 0.58 mol C₆H₁₄ *
= 5.51 mol O₂
As there are more O₂ moles than required (7.81 vs 5.51), O₂ is the reactant in excess. That means that <em>C₆H₁₄ is the limiting reactant</em>.
Now we can <u>calculate how much water can be formed</u>, using <em>the number of moles of the limiting reactant</em>:
- 0.58 mol C₆H₁₄ *
= 4.06 mol H₂O