consider the chemical equation.\n\ncucl₂ + 2nano₃ → cu(no₃)₂ + 2nacl\n\nwhat is the percent yield of nacl if…

Explanation:

Step1: Calculate molar mass of CuCl₂

The molar mass of Cu (copper) is approximately 63.55 g/mol and of Cl (chlorine) is approximately 35.45 g/mol. For CuCl₂, molar mass $M_{CuCl_2}=63.55 + 2\times35.45=134.45$ g/mol.

Step2: Calculate moles of CuCl₂

Given mass of CuCl₂ is $m_{CuCl_2} = 31.0$ g. Moles of CuCl₂, $n_{CuCl_2}=\frac{m_{CuCl_2}}{M_{CuCl_2}}=\frac{31.0}{134.45}\approx0.231$ mol.

Step3: Determine moles of NaCl from stoichiometry

From the balanced chemical equation $CuCl_2 + 2NaNO_3\rightarrow Cu(NO_3)2+2NaCl$, the mole - ratio of $CuCl_2$ to $NaCl$ is 1:2. So moles of NaCl produced theoretically, $n{NaCl - theoretical}=2\times n_{CuCl_2}=2\times0.231 = 0.462$ mol.

Step4: Calculate theoretical mass of NaCl

The molar mass of NaCl is $M_{NaCl}=22.99 + 35.45=58.44$ g/mol. Theoretical mass of NaCl, $m_{NaCl - theoretical}=n_{NaCl - theoretical}\times M_{NaCl}=0.462\times58.44\approx27.0$ g.

Step5: Calculate percent - yield of NaCl

Actual yield of NaCl is $m_{NaCl - actual}=21.2$ g. Percent - yield $=\frac{m_{NaCl - actual}}{m_{NaCl - theoretical}}\times100=\frac{21.2}{27.0}\times100\approx78.5% \approx 78.7%$ (rounding difference).

Answer:

D. 78.7%