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

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: CuCl_2 + NaNO_3 rightarrow Cu(NO_3)_2 + NaCl (b.) If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how many grams of sodium chloride can be formed? (c.) How many grams of copper(II) nitrate can be formed? (d.) If 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction? (e.) Draw the Lewis dot structure for the NO_3^- anion. (f.) The Lewis dot structure for the NO_3^- anion above is one of three resonance structures. Draw the other two resonance structures for the NO_3^- anion in part (e.)

Explanation / Answer

CuCl2 + 2NaNO3 -------> Cu(NO3)2 + 2NaCl

no of moles of CuCl2 = W/G.M.Wt Of CUCl2      = 15/134.5 = 0.11 moles

no of moles of NaNO3 = W/G.M.Wt of NaNO3    =20/85 = 0.235 moles

From the balanced equation

1 mole of CuCl2 react 2 moles of NaNO3

0.11 mole of CuCl2 react with = 2*0.11/1 = 0.22moles of NaNo3 is required

limiting reagent is NaNO3

2 moles of NaNO3 react with CuCl2 to form 2 moles of NaCl

2*85 g of NaNO3 react with CuCl2 to form 2*58.5g of NaCl

20g of NaNO3 react with CuCl2 to form = 2*58.5*20/2*85   = 13.76g of naCl

2 moles of NaNO3 react with CuCl2 to form 1 mole of Cu(NO3)2

2*85g of NaNO3 react with CuCl2 to form 187.56 g of Cu(NO3)2

20 g of NaNO3 react with CuCl2to form = 187.56*20/2*85 = 22.06 g of CU(NO3)2

percentage of NaCl = actual yield*100/theoritical yield

                                     =11.3*100/13.76 = 82.12%

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