This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Step 1: Write and balance the chemical equation for the combustion of butane. Butane (C_4H_10) burns in oxygen (O_2) to produce carbon dioxide (CO_2) and water vapor (H_2O). 2C_4H_10(g) + 13O_2(g) 8CO_2(g) + 10H_2O(g) Step 2: Calculate the molar masses of the reactants and products. • Molar mass of C_4H_10 = (4 × 12.01) + (10 × 1.008) = 48.04 + 10.08 = 58.12 g/mol • Molar mass of O_2 = 2 × 16.00 = 32.00 g/mol • Molar mass of CO_2 = 12.01 + (2 × 16.00) = 12.01 + 32.00 = 44.01 g/mol Step 3: Convert the given masses of reactants to moles. Given: 4.65 g of butane and 40.0 g of oxygen. Moles of C_4H_10: n_C_4H_10 = 4.65 g58.12 g/mol = 0.07999 mol Moles of O_2: n_O_2 = 40.0 g32.00 g/mol = 1.25 mol Step 4: Determine the limiting reactant. Using the stoichiometric coefficients from the balanced equation: For C_4H_10: 0.07999 mol C_4H_10 × 13 mol O_22 mol C_4H_10 = 0.5199 mol O_2 needed. Since we have 1.25 mol O_2, which is more than 0.5199 mol O_2 needed, C_4H_10 is the limiting reactant. O_2 is the excess reactant. Step 5: Calculate the mass of carbon dioxide produced based on the limiting reactant. From the balanced equation, 2 mol C_4H_10 produces 8 mol CO_2. Moles of CO_2 produced: n_CO_2 = 0.07999 mol C_4H_10 × 8 mol CO_22 mol C_4H_10 = 0.31996 mol CO_2 Mass of CO_2 produced: Mass_CO_2 = 0.31996 mol × 44.01 g/mol = 14.08 g The mass of carbon dioxide produced is 14.08 g. Step 6: Calculate the mass of the excess reactant (O_2) remaining. Moles of O_2 consumed: n_O_2, consumed = 0.07999 mol C_4H_10 × 13 mol O_22 mol C_4H_10 = 0.5199 mol O_2 Moles of O_2 remaining: n_O_2, remaining = 1.25 mol (initial) - 0.5199 mol (consumed) = 0.7301 mol O_2 Mass of O_2 remaining: Mass_O_2, remaining = 0.7301 mol × 32.00 g/mol = 23.36 g The mass of excess reactant remaining is 23.36 g. Drop the next question.