A0.5674 g piece of copper is added to 10.00 ml; of 16 m hno3, producing 0.8024 g of copper (ii) nitrate. what is the precent yield of the reaction?

The percent yield of the reaction is 48.05%

Explanation:

We have the following balanced reaction:

Cu + 2HNO3 = Cu(NO3)2 + H2

The moles of copper is equal to:

moles Cu = 0.5674 g/63.54 g/mol = 0.0089 moles

The moles of HNO3 is equal to:

moles HNO3 = 10 mL * (1 L/1000 mL) * 16 M = 0.16 moles

The mass of Cu produced is equal to:

mass of Cu(NO3)2 = moles of Cu(NO3)2 * molar mass Cu(NO3)2

mass of Cu(NO3)2 = 0.0089 mol * 187.56 g/mol = 1.67 g

The percent yield is equal to:

percent yield = (0.8024/1.67)*100 = 48.05%

50.23 i believe is correct but don't quote me on that.