Answer to Question #173645 in General Chemistry for Sophia

Question

When copper (II) chloride reacts with sodium nitrate, copper (II) nitrate and sodium chloride are formed

A. Write the balanced equation for the reaction given above:

_CuCI2+_NaNO3 = _Cu(NO3)2+_ NaCI

B. If 15 grams of copper (II) chloride react with 20.0 grams of sodium nitrate, how much sodium chloride can be formed?

Grams of NaCI(15.0g CuCI) I =

Grams of NaCI( 20.0g NaNO3)=

C. What is the limiting Reagent for the reaction?(chemical formula)

D. How many grams of copper (II) nitrate If formed? (Numbers only)

E. How much of the excess reagent is left over in this reaction? (Question D) numbers only

F. If 11.3 grams of sodium chloride are formed in the reaction described in problem B, what is the percent yield of this reaction?

Answer 1

$a) CuCI_2+2NaNO_3=Cu(NO3)_2+2NaCI$

b) $m(NaCl)=m(CuCI2)×2Mr(NaCl)/Mr(CuCI2)=15×2×(94/135) =20.89 g$

c) $CuCI_2$ - limiting reactant;

$NaNO_3$ - excess reactant

d)$mleft(NaNO_3)=20 - (15×170/135)=1.1 g$

e) $Cu(NO3)_2=0.92×15×188/135=19.22 g$

f) $65.7\%$