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If 15 grams of copper (ii) chloride react with 20 grams of sodium nitrate, how many grams of sodium chloride can be formed?

User Kgm
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Answer: approximately 6.53 grams of sodium chloride can be formed in this reaction.

Explanation: The balanced equation for the reaction is:

2 CuCl2 + 2 NaNO3 → Cu(NO3)2 + 2 NaCl

According to the balanced equation, 2 moles of CuCl2 react with 2 moles of NaNO3 to produce 1 mole of Cu(NO3)2 and 2 moles of NaCl.

Next, we can calculate the moles of CuCl2 and NaNO3 provided in the problem:

Mass of CuCl2 = 15 grams

Molar mass of CuCl2 = 63.5 + (2 x 35.45) = 134.4 g/mol

Moles of CuCl2 = Mass of CuCl2 / Molar mass of CuCl2 = 15 g / 134.4 g/mol ≈ 0.1119 mol

Mass of NaNO3 = 20 grams

Molar mass of NaNO3 = 22.99 + 14.01 + (3 x 16.00) = 85 g/mol

Moles of NaNO3 = Mass of NaNO3 / Molar mass of NaNO3 = 20 g / 85 g/mol ≈ 0.235 mol

Since the mole ratio between CuCl2 and NaCl is 2:2, the limiting reactant in this case is CuCl2 because it is present in lesser amount (0.1119 mol) compared to NaNO3 (0.235 mol).

From the balanced equation, we can see that 2 moles of CuCl2 react to produce 2 moles of NaCl. Therefore, the theoretical yield of NaCl is also 0.1119 mol.

Finally, we can calculate the mass of NaCl formed:

Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol

Mass of NaCl = Moles of NaCl x Molar mass of NaCl = 0.1119 mol x 58.44 g/mol ≈ 6.53 grams

So, approximately 6.53 grams of sodium chloride can be formed in this reaction.

User Racket Noob
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