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?

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joseaaronlara joseaaronlara · Answer 1 · 2019-10-15T02:34:57+02:00

Answer:

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 %