Question

2.00g of Copper wire was added to 50.0 mL of a 0.050M solution of Silver Nitrate. The reaction

Answer 1

Answer:

See explanation

Explanation:

Now, the equation of the reaction is;

Cu(s) + 2AgNO3(aq) --------> Cu(NO3)3 (aq) + 2Ag(s)

Number of moles of silver precipitate = 0.24/108 g/mol= 0.0022 moles

Number of moles of AgNO3 = 50/1000 * 0.050 = 0.0025 Moles

If 2 moles of AgNO3 yields 2 moles of Ag

Then  0.0025 moles of AgNO3 yields  0.0025 * 2/2 = 0.0025 moles of Ag

Number of moles of Cu = 2.00g/63.5 g/mol = 0.03 moles

If 1 mole of Cu yields 2 moles of Ag

0.03 moles of Cu yields  0.03 moles * 2/1 = 0.06 moles of Ag

Hence AgNO3 is the limiting reactant

Theoretical yield of Ag = 0.0025 moles of Ag * 108 = 0.27 g

Actual yield of Ag = 0.24 g

Percent yield = actual yield/theoretical yield * 100

Percent yield = 0.24/0.27 * 100

Percent yield = 88.8%