Consider the chemical equation CuCl_(2)+2NaNO_(3)arrow Cu(NO_(3))_(2)+2NaCl What is the percent yield of NaCl if 31 og of CuCl_(2) reacts with excess NaNO_(3) to produce 2129 of NaCl? Use % Yield=(Actualyield)/(Theoreticalyreld)times 100 49.7% 58.4% 63.6% 78.7%

To calculate the percent yield of NaCl, we first need to determine the theoretical yield of NaCl.<br /><br />Given information:<br />- 31 g of CuCl2 reacts with excess NaNO3 to produce 2129 g of NaCl.<br /><br />Step 1: Calculate the molar mass of CuCl2.<br />Molar mass of CuCl2 = 63.55 (Cu) + 2 × 35.45 (Cl) = 134.45 g/mol<br /><br />Step 2: Calculate the number of moles of CuCl2.<br />Number of moles of CuCl2 = 31 g / 134.45 g/mol = 0.230 mol<br /><br />Step 3: Calculate the theoretical yield of NaCl.<br />According to the balanced chemical equation, 1 mol of CuCl2 produces 2 mol of NaCl.<br />Theoretical yield of NaCl = 0.230 mol × 2 × 58.44 g/mol = 26.9 g<br /><br />Step 4: Calculate the percent yield of NaCl.<br />Percent yield = (Actual yield / Theoretical yield) × 100<br />Percent yield = (2129 g / 26.9 g) × 100 = 78.7%<br /><br />Therefore, the correct answer is $78.7\%$.