Write and balance the chemical equation for the combustion of butane.

Here's how to solve problem 3.16B: 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 (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 involved. • 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. • Moles of C_4H_10 = 4.65 g58.12 g/mol = 0.07999 mol • Moles of O_2 = 10.0 g32.00 g/mol = 0.3125 mol Step 4: Determine the limiting reactant. Using the stoichiometric ratio from the balanced equation (2 mol C_4H_10 reacts with 13 mol O_2): • If C_4H_10 is limiting, moles of O_2 needed = 0.07999 mol C_4H_10 × 13 mol O_22 mol C_4H_10 = 0.5199 mol O_2. Since we only have 0.3125 mol O_2, oxygen is the limiting reactant. Step 5: Calculate the mass of carbon dioxide produced using the limiting reactant (O_2). From the balanced equation, 13 mol O_2 produces 8 mol CO_2. • Moles of CO_2 produced = 0.3125 mol O_2 × 8 mol CO_213 mol O_2 = 0.1923 mol CO_2 • Mass of CO_2 produced = 0.1923 mol × 44.01 g/mol = 8.463 g Step 6: Calculate the mass of the excess reactant (C_4H_10) remaining. First, find the moles of C_4H_10 that reacted with the limiting reactant. • Moles of C_4H_10 reacted = 0.3125 mol O_2 × 2 mol C_4H_1013 mol O_2 = 0.04808 mol C_4H_10 • Mass of C_4H_10 reacted = 0.04808 mol × 58.12 g/mol = 2.795 g • Mass of C_4H_10 remaining = Initial mass - Mass reacted = 4.65 g - 2.795 g = 1.855 g Rounding to three significant figures: • Mass of carbon dioxide produced: 8.46 g • Mass of excess reactant (C_4H_10) remaining: 1.86 g That's 2 down. 3 left today — send the next one.