0.22 mol NaCl(58.44g NaCl/1 mole NaCl)=12.86g
15g copper(II) chloride can form 12.86g sodium chloride.
let us find the limiting reactant.
Earlier we calculated that the moles that can be formed by 15g copper(II) chloride is 12.86g. Now let us find how many moles it will take for 20g Sodium Nitrate.
0.24 mol NaCl(58.44g NaCl/1 mole NaCl)=14.03g NaCl.
The limiting reactant is the copper(II) chloride.
The excess reagent that is left is 1.17.
Actual yield:11.3 theoretical yield:12.86
Percent yield: 11.7/12.86 x 100=90.97%