What is the percent yield of nacl if 31.0 g of cucl reacts with excess nano3 to produce 21.2 g of nacl

Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100  

The Actual Yield is given in the question as 21.2 g of NaCl.  However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.

Balanced Equation:   CuCl + NaNO₃    →    NaCl + CuNO₃

Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
                         =  31.0 g ÷ (63.5 + 35.5)g/mol
                         = 0.31 mol

the mole ratio of CuCl to NaCl is 1  :  1,
∴ if moles of CuCl = 0.31  mol,

then moles of NaCl = 0.31 mol

Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
                                 =  0.31 mol × (23 + 35.5) g/mol
                                 =  18.32 g

⇒ the THEORETICAL Yield of NaCl, in this case, is 18.32 g.

Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100  

⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100  
                                                = 115.7 %

NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.

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