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For the combustion of ethanol (C2H5OH), we know that ethanol reacts with oxygen (O2) to produce water (H2O) and carbon dioxide (CO2). The balanced chemical equation for this reaction is: C2H5OH + 3O2 -> 2CO2 + 3H2O

Using the molar masses of ethanol (C2H5OH) and oxygen (O2), we can convert the given mass of each reactant to moles: Molar mass of ethanol (C2H5OH) = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.07 g/mol Moles of ethanol = \( \frac{46.0 \mathrm{~g}}{46.07 \mathrm{~g/mol}} \) = 1.0 mol Molar mass of oxygen (O2) = 2(16.00 g/mol) = 32.00 g/mol Moles of oxygen = \( \frac{96.0 \mathrm{~g}}{32.00 \mathrm{~g/mol}} \) = 3.0 mol

Using the balanced chemical equation, we can determine the stoichiometric ratio of moles of ethanol to moles of oxygen. We will then use this ratio to determine the limiting reactant: Stoichiometric ratio: 1 mol C2H5OH : 3 mol O2 For 1.0 mol of ethanol, theoretically, we would require 3.0 mol of oxygen. Since we have exactly this amount, both reactants are consumed completely, and there is no limiting reactant.

We are given the mass of water produced (54.0 g) and need to calculate the mass of carbon dioxide produced. To do this, we can use the stoichiometry of the balanced chemical equation: Stoichiometric ratio: 2 mol CO2 : 3 mol H2O First, we need to convert the mass of water produced to moles: Molar mass of water (H2O) = 2(1.01 g/mol) + 1(16.00 g/mol) = 18.02 g/mol Moles of water = \( \frac{54.0 \mathrm{~g}}{18.02 \mathrm{~g/mol}} \) = 3.0 mol Using the stoichiometric ratio, we can determine the moles of carbon dioxide: Moles of CO2 = \( \frac{2 \mathrm{~mol \, CO2}}{3 \mathrm{~mol \, H2O}} \) × 3.0 mol H2O = 2.0 mol CO2 Now, we can convert the moles of carbon dioxide to mass: Molar mass of carbon dioxide (CO2) = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol Mass of CO2 = 2.0 mol CO2 × 44.01 g/mol = 88.0 g The mass of carbon dioxide produced is 88.0 g.