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: b) if 45 grams of copper (ii) chloride react with 25 grams of sodium nitrate, how much c) what is the limiting reagent for the reaction? d) how many grams of copper(ii) nitrate is formed? e) how much of the excess reagent is left over in this reaction? f) if 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction?

The answer to your question is:

Explanation:

a)    CuCl₂   +  2NaNO₃    ⇒     Cu(NO₃)₂   + 2NaCl

b)

45 g CuCl₂

25 g NaNO₃                 Is not complete the question

c) MW CuCl₂ = 64 + 71 = 135 g

   MW NaNO₃ = 23 + 14 + 48 = 85 x 2 = 170 g

theoretical Ratio      170 / 135 = 1.26

experimental ratio   25/ 45 = 0.56

Limiting reactant = NaNO₃

d)                    170 g of NaNO₃  126 g of Cu(NO₃)₂

                    25 g                    x

                  x = (25 x 126)/170

                  x = 18.52 g of Cu(NO₃)₂

e)                170 g of NaNO₃    135 g of CuCl₂

                    25 g of NaNO₃    x

                   x = 19.85 g of CuCl₂

Excess = 45 -  19.85 = 25.15 g of CuCl₂

f) 11.3 g of NaCl

Theoretical

                    170 g of NaNO₃  117 g of NaCl

                      25 g                   x

                     x = (25 x 117) / 170

                     x = 17.2 g of NaCl

yield = 11.3/17.2 x 100

yield = 65.7 %

where are the choice options?

i think its 100.

hope this !