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. Write the balanced equation for the reaction given above: If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate, how many grams of sodium chloride can be formed? How many grams of copper(II) nitrate can be formed? If 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction? Draw the Lewis dot structure for the NO_3 anion. 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

A) Ans: To balance a chemical equation we need to look at both the sides of the reaction and decide the stoichiometric coefficient.

Here we see that product has got 2NO3 so the reactant too should have 2NO3. Here we should make the stoichiometry coefficient of the reactant containing NO3 as 2 and after this we make the stoichiometry coefficient of the NaCl as 2 as well for the reaction to balance.

So the balanced reaction is

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

B) Ans: For finding the weight of compound formed in a reaction we need to look at the limiting reagent. The limiting agent is the reactant that gets consumed totally (first). So we need to find the numbers of moles of the reactant and look at their stoichiometric coefficient for deciding the limiting reagent.

No. of moles of 15g Cu(II) Chloride = 15/ 134.45 = 0.112 moles (Mol. wt of CuCl2 = 134.45g)

No. of moles of 20g sodium nitrate = 20/84.99 = 0.235 moles (Mol. wt of NaNO3 = 84.99g )

The balanced chemical equation says that one mole of CuCl2 reacts with 2 moles of NaNO3 to the balanced products.

The no. of moles of NaNO3 compared to CuCl2 = 0.235/0.112 = 2.1

Hence the limiting agent will be CuCl2.

The balance chemical equation says that NaCl formed will be 2 times the mole of CuCl2

Hence, No. of moles of NaCl formed = 2* 0.112 = 0.224

Weight of NaCl formed = 0.224 * 58.44 = 13.09 g (Mol. wt of NaCl = 58.44 g)

C) Ans: The balance chemical equation says that Cu(NO3)2 formed will be equal to the mole of CuCl2

So, Weight of Cu(NO3)2 formed = 0.112* 187.55 = 21.00 g (Mol. wt of Cu(NO3)2 = 187.55 g)

Thank You

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