Question

What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to produce 21.2 g of NaCl?

Answer 1

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!