Mishieeidn
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CuCl(subscript 2) + 2NaNO(subscript 3) --> Cu(NO(subscript 3)2 + 2NACl
a. if 15 grams of copper(ll) chloride react with 20 grams of sodium nitrate how much sodium chloride can be formed?

b. What is the name of the limiting reagent?
c.How much of the excess reagent is left over in this reaction?
D. if 11.3 grams of sodium chloride are formed in the reaction what is the percent yield of this reaction ?

pls help me answer these questions . longing for your help thank you

Sagot :

Jers15idn
[tex]CuCl_{2}+2NaNO_{3} --->Cu(NO_{3})_{2}+2NaCl [/tex]

a.Mole relationship:[tex]1 mole CuCl_{2}/2 mol NaCl[/tex]
[tex]15g CuCl_{2}(1 mole CuCl_{2}/134.45g CuCl_{2})=0.11 mol CuCl_{2}[/tex]
[tex]0.11 mol CuCl_{2}(2 mol NaCl/1 mol CuCl_{2})=0.22 mol NaCl[/tex]
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.


[tex]20g NaNO_{3}(1 mole NaNO_{3}/85g NaNO_{3})=0.24 moles NaNO_{3}[/tex]
[tex]0.24 mol NaNO_{3}(2 mol NaNO_{3}/2 mol NaCl)=0.24 mol NaCl[/tex]
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%



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