consider the chemical equation. cucl2 + 2nano3 → cu(no3)2 + 2nacl what is the percent yield of nacl if 31.0…

consider the chemical equation. cucl2 + 2nano3 → cu(no3)2 + 2nacl what is the percent yield of nacl if 31.0 g of cucl2 reacts with excess nano3 to produce 21.2 g of nacl? use %yield = actual yield / theoretical yield × 100. 49.7% 58.4% 63.6% 78.7%
Answer
Explanation:
Step1: Calculate molar mass of CuCl₂
The molar mass of Cu (copper) is approximately 63.55 g/mol and Cl (chlorine) is approximately 35.45 g/mol. For CuCl₂, molar - mass = 63.55+(2×35.45)=134.45 g/mol.
Step2: Calculate moles of CuCl₂
Moles of CuCl₂ = $\frac{mass}{molar - mass}=\frac{31.0\ g}{134.45\ g/mol}\approx0.231\ mol$.
Step3: Determine moles of NaCl from stoichiometry
From the balanced chemical equation CuCl₂ + 2NaNO₃ → Cu(NO₃)₂+2NaCl, the mole - ratio of CuCl₂ to NaCl is 1:2. So moles of NaCl produced theoretically = 2×0.231 mol = 0.462 mol.
Step4: Calculate theoretical mass of NaCl
The molar mass of NaCl is 22.99 + 35.45=58.44 g/mol. Theoretical mass of NaCl = moles×molar - mass = 0.462 mol×58.44 g/mol≈27.0 g.
Step5: Calculate percent yield
Percent yield = $\frac{Actual\ yield}{Theoretical\ yield}\times100=\frac{21.2\ g}{27.0\ g}\times100\approx78.7%$.
Answer:
78.7%