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noname [10]
3 years ago
6

2 C6H14 + 19 O2 --> 12 CO2 + 14 H2O

Chemistry
1 answer:
Rashid [163]3 years ago
5 0

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 * \frac{19molO_2}{2molC_6H_{14}} = 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 * \frac{14molH_2O}{2molC_6H_{14}} = 4.06 mol H₂O
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