Question

a) Balance the equation for the reaction of iron metal with silver (I) nitrate to form iron (II) nitrate
and silver metal.
b) Using the equation from part a, determine the mass of silver metal that can be made if the reaction
is performed using 67.3 g iron metal and 75.3 g silver (I) nitrate. HINT: This is a limiting reactant
problem!
c) If the actual yield of silver metal is 45.1 g when the reaction is performed, what is the percentage
of the theoretical yield?

Answer 1

a)

this is a single replacement reaction

Fe(s) + 2AgNO3(aq) ---> Fe(NO3)2 + 2Ag(s)

b)

mass of Ag from

m = 67.3 g of Iron

mol = mass/MW = 67.3/55.5 = 1.2126 mol of Fe

and

m = 75.3 g of AgNO3

MW = 169.87

mol = 75.3/169.87 = 0.443280 mol of AgNO3

then

since ratio is 2:2

0.443280 mol of AgNO3 produces --> 0.443280 mol of Ag

mass = mool*MW = 107.8682 *0.443280 = 47.8158g of Ag

c)

if real yield = 45.1 then

%yield = real/theoretical * 100 = 45.1/47.8158 *100 = 94.32028%