The percent yield of copper hydroxide is 84%
<h3>Stoichiometry</h3>From the question, we are to determine the percent yield of copper hydroxide
First, we will determine the theoretical mass
From the given balanced chemical equation, we have
Cu(NO₃)₂ + 2NaOH -- Cu(OH)₂ + 2NaNO₃
This means,
1 mole of copper(II) nitrate reacts with 2 moles of sodium hydroxide to produce 1 mole of copper hydroxide
Therefore,
0.05 mole of copper(II) nitrate reacts with 0.1 mole of sodium hydroxide to produce 0.05 mole of copper hydroxide
The theoretical number of moles of copper hydroxide that is produced is 0.05 mole
Now, for the theoretical mass
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of copper hydroxide = 97.56 g/mol
Then,
Theoretical mass = 0.05 × 97.56
Theoretical mass of copper of hydroxide produced is = 4.878 g
Now, for the percent yield of copper hydroxide
Percent yield is given by the formula,
Then,
Hence, the percent yield of copper hydroxide is 84%.
Learn more on Stoichiometry here: brainly.com/question/9372758