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 for the reaction. Calcium (Ca) reacts with Nitrogen gas (N₂) to form Calcium nitride (Ca₃N₂). 3Ca(s) + N_2(g) → Ca_3N_2(s) Step 2: Calculate the molar masses of the reactants and the product. • Molar mass of Ca = 40.08 g/mol • Molar mass of N₂ = 2 × 14.01 g/mol = 28.02 g/mol • Molar mass of Ca₃N₂ = (3 × 40.08 g/mol) + (2 × 14.01 g/mol) = 120.24 g/mol + 28.02 g/mol = 148.26 g/mol Step 3: Convert the given masses of reactants to moles. • Moles of Ca = 56.6 g40.08 g/mol = 1.4122 mol Ca • Moles of N₂ = 30.5 g28.02 g/mol = 1.0885 mol N_2 Step 4: Determine the limiting reactant. Using the stoichiometric ratio from the balanced equation (3 mol Ca : 1 mol N_2): • To react all 1.4122 mol Ca, we need: 1.4122 mol Ca × 1 mol N_23 mol Ca = 0.4707 mol N_2 Since we have 1.0885 mol N_2, which is more than 0.4707 mol N_2, nitrogen is in excess. Therefore, Calcium (Ca) is the limiting reactant. Step 5: Calculate the theoretical yield of Calcium nitride (Ca₃N₂). The theoretical yield is based on the limiting reactant, Ca. From the balanced equation, 3 mol Ca produces 1 mol Ca_3N_2. • Moles of Ca₃N₂ produced = 1.4122 mol Ca × 1 mol Ca_3N_23 mol Ca = 0.4707 mol Ca_3N_2 • Mass of Ca₃N₂ (theoretical) = 0.4707 mol × 148.26 g/mol = 69.79 g Ca_3N_2 Step 6: Calculate the actual mass of Calcium nitride formed using the given percent yield. The percent yield is 93.0\%. Actual yield = Theoretical yield × Percent yield100\% Actual yield = 69.79 g × (93.0)/(100) = 69.79 g × 0.930 = 64.9047 g Rounding to three significant figures, the actual mass of calcium nitride formed is 64.9 g. The mass of Calcium nitride formed is 64.9 g. Send me the next one 📸