2 C6H14 + 19 O2 –> 12 CO2 + 14 H2O

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

Given 250 g of Oxygen and 50 grams of CoH14 what is the maximum amount of water that can be formed? • Box 1 = number • Box 2 = units Box 3 = Substance​

Answers

4.06 mol H₂O

Explanation:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

First we convert the given masses of reactants into moles, using their respective molar masses:

250 g O₂ ÷ 32 g/mol = 7.81 mol O₂50 g C₆H₁₄ ÷ 86 g/mol = 0.58 mol C₆H₁₄

Now we calculate how many O₂ moles would react completely with 0.58 C₆H₁₄ moles, using the stoichiometric coefficients of the reaction:

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 C₆H₁₄ is the limiting reactant.

Now we can calculate how much water can be formed, using the number of moles of the limiting reactant:

0.58 mol C₆H₁₄ * frac{14molH_2O}{2molC_6H_{14}} = 4.06 mol H₂O

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