Question

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

produced solid silver and aqueous copper II Nitrate. The silver precipitate was collected and
dried. It was determined that the mass of the silver precipitate was 0.24 g. Use this information
to determine the theoretical yield, limiting reactant, and percent yield.

Answer 1

See explanation

Step-by-step 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%