A copper bar with a mass of 12.340 g is dipped into 255 mL of a solution contain

Question

A copper bar with a mass of 12.340 g is dipped into 255 mL of a solution containing 0.125 M silver ions. When the reaction that occurs finally ceases, what will be the mass of unreacted copper in the bar? What will be the mass of silver metal that "plates out" of solution (that is, what mass of silver metal is produced by the reaction)? (hint: the oxidation state of copper in aqueous solution is +2) A. 10.31 g of unreacted copper in the bar; 1.72 g of silver metal produced B. 11.33 g of unreacted copper in the bar; 1.72 g of silver metal produced C. no copper bar remains; 41.9 g of silver metal produced D. 10.31 g of unreacted copper in the bar; 3.44 g of silver metal produced E. 11.33 g of unreacted coppper in the bar; 3.44 g of silver metal produced

Explanation / Answer

The amount of AgNO3 present in reaction is n1 = 0.125 * 0.255 mole = 31.87 mmole.
The reaction is: Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag | . If 1 mmole Cu uses 2mmole AgNO3, then n2 = n1 /2 = 15.935mmole Cu will be dissolved, so m = n*M = 15.935*10^-3 * 64 = 1.02g of copper.That means delta m = 12.340 - 1.02 = 11.32g.

B. 11.33 g of unreacted copper in the bar; 1.72 g of silver metal produced

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