Copper dissolves in nitric acid, producing copper (ll) nitrate. After reaction with sodium hydroxide and heating, copper nitrate changes to copper (ll) oxide. Calculate the percent yield of this process if 1.235 g of copper produced 1.452 g of copper oxide.
Convert 1.235 g Cu to mols Cu.
Convert mols Cu to mols Cu(II) nitrate and to mols Cu(II) oxide.
Convert mols Cu(II) oxide to grams Cu(II) oxide. This is the theoretical yield in grams.
Then % yield = (1.452 g/theoretical yield)*100 =
To calculate the percent yield of a reaction, you need to compare the actual yield with the theoretical yield. The theoretical yield is the maximum amount of product that could be obtained under ideal conditions, while the actual yield is the amount that is actually obtained in the experiment.
To determine the theoretical yield, you need to calculate the amount of copper (ll) oxide that would be formed from the given amount of copper. The balanced chemical equation for the reaction is:
2Cu + 4HNO3 -> 2Cu(NO3)2 + 2NO2 + 2H2O
3Cu(NO3)2 + 2NaOH -> Cu3O + 2NaNO3 + 2H2O
From the balanced equation, you can see that the stoichiometric ratio between copper (ll) and copper (ll) oxide is 2:1. This means that for every 2 moles of copper reacted, you would expect to obtain 1 mole of copper (ll) oxide.
Now, let's calculate the moles of copper used and the moles of copper (ll) oxide obtained:
Molar mass of copper (Cu) = 63.55 g/mol
Molar mass of copper (ll) oxide (Cu3O) = 79.55 g/mol
Moles of copper (Cu):
1.235 g / (63.55 g/mol) = 0.0194 mol
Moles of copper (ll) oxide (Cu3O):
1.452 g / (79.55 g/mol) = 0.0182 mol
Now you can determine the theoretical yield using the ratio mentioned earlier:
Theoretical yield = (0.0182 mol Cu3O / 0.0194 mol Cu) * 1.235 g Cu = 1.154 g Cu3O
Finally, you can calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (1.452 g / 1.154 g) * 100 = 125.9%
Therefore, the percent yield of the process is 125.9%.