Analysis - Limiting Reagent, Theoretical Yield and Percentage Yield

Based on our observation, these are the amount of:

Silver nitrate: 1.045 g

Copper: 0.762 g

Resulting Copper nitrate: 065 g

Resulting Silver: 1.837 g

To find the limiting reagent we have to balance the equation first:

2 AgNO₃ + Cu --> Cu(NO₃)₂ + 2 Ag

Now we have to find the mol of each resulting elements:

nAgNO₃ = 1.045 g AgNO₃ x 1 mol AgNO₃/(107.8682 + 14.0067 + 2(15.9994)) g AgNO₃ = 0.00615165 mol AgNO₃

nCu = 0.762 g Cu x 1 mol Cu/63.546 g Cu = 0.011991 mol Cu

For the last step, we need to make the mol equal:

nCu = 0.011991 mol Cu x 2 mol AgNO₃/1 mol Cu = 0.023982 mol AgNO₃

:: From the calculation we can see that silver nitrate has less mol than copper, therefore silver nitrate is the limiting reagent. We can find the theoretical yield of the experiment by calculation using the limiting reagent:

MAg = 1.045 AgNO3 x 1 mol AgNO3/169.8731 g AgNO3 x 2 mol Ag/2 mol AgNO3 x 107.8682 g Ag/1 mol AgNO3 = 0.66 g Ag

MCu(NO₃)₂ = 1.045 AgNO₃ x 1 mol AgNO₃/169.8731 g AgNO₃ x 1 mol Cu(NO₃)₂/2 mol Ag(NO₃)₂ x 187.5558 g Cu(NO₃)₂/1 mol Cu(NO₃)₂ = 0.57 g Cu(NO₃)₂

:: The calculation shows that the theoretical yield of the resulting silver is 0.66 g and the theoretical yield of the resulting copper nitrate is 0.57 g. If we found the theoretical yield we can found the percentage yield:

% of Ag = 1.84g Ag/066 g Ag x 100% = 279%

% of Cu(NO₃)₂ = 0.65 g Cu(NO₃)₂/ 0.57 g Cu(NO₃)₂ x 100% = 114%