Consider the chemical equation. cucl2 + 2nano3 mc023-1.jpg 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?

CuCl2 + 2NaNO3 >  Cu(NO3)2 + 2NaCl

using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g  of CuCl2  produces  2(22.98 + 35.45) g of NaCl
    134.44  g 116.86 g
       31.0 g 31.0 * 116.86 /134.44=26.95g  
 
 So percentage yield is 21.2* 100 / 26.95    =  78.7%  to nearest tenth                                                                                                              

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