What is the percent yield of nacl if 31.0 g of cucl2 reacts with excess nano3 to produce 21.2 g of nacl?

First you need to find the balanced chemical formula.
CuCl₂+2NaNO₃→2NaCl+Cu(NO₃)₂

Then you can find out how much NaCl 31.0g of CuCl₂ should produce using stoichiometry.  Divide 31.0g by the molar mass of CuCl₂ (134.446g/mol) to get 0.2306mol CuCl₂.  Than multiply 0.2306mol CuCl₂ by 2 to get 0.4612mol NaCl.  Than multiply 0.4612mol by the molar mass of NaCl (58.45g/mol) to get 26.95g of NaCl.
that means that 100% yield would give you 26.95g of NaCl so to find percent yield divide 21.2 by 26.95 to get 0.7867 which is 78.7% yield

Therefore the answer is 78.7% yield.
I hope this helps.  Let me know if anything is unclear

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