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If 27.3g of C8H18 is combusted, according to the equation below, what mass of water is produced?

2C8H18 + 25O2 --> 16CO2 + 18 H2O,

molar masses: C8H18 = 114.3 g/mol O2 = 32.0 g/mol CO2 = 44.0 g/mol H2O = 18.0 g/mol

Respuesta :

maacastrobr

Answer:

38.7 g

Explanation:

  • 2C₈H₁₈ + 25O₂ → 16CO₂ + 18 H₂O

First we convert 27.3 grams of C₈H₁₈ into moles, using the given molar mass:

  • 27.3 g C₈H₁₈ ÷ 114.3 g/mol = 0.239 mol C₈H₁₈

Then we convert 0.239 moles of C₈H₁₈ into moles of H₂O, using the stoichiometric coefficients of the reaction:

  • 0.239 mol C₈H₁₈ * [tex]\frac{18molH_2O}{2molC_8H_{18}}[/tex] = 2.151 mol H₂O

Finally we convert 2.151 moles of H₂O into grams, using water's molar mass:

  • 2.151 mol H₂O * 18.0 g/mol = 38.7 g

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