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 the balanced chemical equation. Aluminum reacts with sulfur to form aluminum sulfide. Aluminum typically forms a Al^3+ ion and sulfur forms a S^2- ion, so the compound is Al_2S_3. 2Al(s) + 3S(s) → Al_2S_3(s) Step 2: Calculate the molar masses of the reactants and product. • Molar mass of Al = 26.98 g/mol • Molar mass of S = 32.07 g/mol • Molar mass of Al_2S_3 = (2 × 26.98 g/mol) + (3 × 32.07 g/mol) = 53.96 g/mol + 96.21 g/mol = 150.17 g/mol Step 3: Convert the given masses of reactants to moles. • Moles of Al = 10.0 g26.98 g/mol = 0.3706 mol • Moles of S = 15.0 g32.07 g/mol = 0.4677 mol Step 4: Determine the limiting reactant. To find the limiting reactant, divide the moles of each reactant by its stoichiometric coefficient from the balanced equation: • For Al: 0.3706 mol2 = 0.1853 • For S: 0.4677 mol3 = 0.1559 Since 0.1559 is less than 0.1853, Sulfur (S) is the limiting reactant, and Aluminum (Al) is the excess reactant. Step 5: Calculate the mass of solid Aluminum Sulfide (Al_2S_3) produced. The amount of product is determined by the limiting reactant (Sulfur). From the balanced equation, 3 moles of S produce 1 mole of Al_2S_3. • Moles of Al_2S_3 produced = 0.4677 mol S × 1 mol Al_2S_33 mol S = 0.1559 mol Al_2S_3 • Mass of Al_2S_3 = 0.1559 mol × 150.17 g/mol = 23.41 g Rounding to three significant figures, the mass of Al_2S_3 produced is 23.4 g. Step 6: Calculate the mass of the non-limiting reactant (Aluminum) in excess. First, calculate the moles of Al consumed by the limiting reactant (S). From the balanced equation, 3 moles of S react with 2 moles of Al. • Moles of Al consumed = 0.4677 mol S × 2 mol Al3 mol S = 0.3118 mol Al • Mass of Al consumed = 0.3118 mol × 26.98 g/mol = 8.413 g Al • Mass of Al in excess = Initial mass of Al - Mass of Al consumed • Mass of Al in excess = 10.0 g - 8.413 g = 1.587 g Rounding to three significant figures, the mass of Aluminum in excess is 1.59 g. Send me the next one 📸