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

Number of moles  CuCl₂ :

mass solute / molar mass

31.0 / 134.4520  => 0.230 moles of CuCl₂

Mole ratio :

CuCl₂ + 2 NaNO₃ = 2 NaCl + Cu(NO₃)₂ 

1 mole CuCl₂ 2 moles NaCl
 0.230 moles  ? moles NaCl

moles NaCl = 0.230 x 2 / 1

moles NaCl = 0.46 / 1

= 0.46  moles of NaCl ( molar mass 58.44 g/mol )

Mass of NaCl:

1 mole 58.44 g
0.46 moles ( mass of NaCl in theory )

mass of NaCl = 0.46 x 58.44 / 1

mass of NaCl = 26.8824 / 1

= 26.8824 g of NaCl

yield of NaCl :

NaCl produce / NaCl in theory  x 100 %

( 21.2 / 26.8824 ) x 100 

( 0.788 ) x 100 => 78.8 %

hope this helps!

The answer is nitrogen-containing bases