Question

If 10.28 g of Cu(NO3)2 are obtained from the reaction of 4.54 g of Cu with excess HNO3, what is the percent yield of the reaction? (show workings for full grade)

Answer 1

The percent yield of the reaction : 77.2%

Further explanation

Percent yield is the comparison of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

mol of Cu(MW=63.55 g/mol) :

`\tt (4.54)/(63.55)=0.071`

mol ratio Cu : Cu(NO₃)₂ = 1 : 1, so mol Cu(NO₃)₂ :

`\tt (1)/(1)* 0.071=0.071`

mass of Cu(NO₃)₂ (MW=187,56 g/mol) :

`\tt mass=mol* MW\\\\mass=0.071* 187,56 g/mol\\\\mass=13.317~g\rightarrow theoretical~yield`

10.28 g of Cu(NO₃)₂ are obtained from the reaction⇒actual yield

The percent yield :

`\tt \%yield=(actual)/(theoretical)* 100\%\\\\\%yield=(10.28)/(13.317)* 100\%\\\\\%yield=77.2\%`