Answer to Question #181065 in General Chemistry for lala

Question

1. Eighty grams of Silver was obtained from one hundred and forty grams of Silver nitrate. The Silver metal is prepared by reducing its nitrate. The chemical equation of the reaction is:

Cu(s) + AgNO3(aq) = Cu(NO3)2(aq) + Ag(s)

a. Balance the chemical equation;

b. Identify the limiting reactant and the excess reactant if applicable;

c. Compute for the theoretical yield; d. Determine the percent yield of the reaction;

e. Calculate the percent error and

f. Compute for the excess amount of the excess reactant if applicable.

g. Show the complete solutions for your answers in a separate paper.

Answer 1

Balanced

Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

Moles of silver nitrate = $\frac{140}{170}=0.82 moles$

Moles of Ag = 0.82 moles

There is no limiting or excess reactant

Theoretical yield = 0.82 × 108 = 88.56 g

Percent yield = $\frac{80}{88.56}×100=90.3$ %