Question

Britney added 0.05 moles of copper(II) nitrate solution to 0.1 moles of sodium hydroxide solution and

Answer 1

The percent yield of copper hydroxide is 84%

Stoichiometry

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,

$Percent\ yield = \frac{Actual\ yield}{Theoretical\ yield} \times 100\%$

Then,

$Percent\ yield\ of\ copper\ hydroxide= \frac{4.1}{4.878}\times 100\%$

$Percent\ yield\ of\ copper\ hydroxide= 84\%$

Hence, the percent yield of copper hydroxide is 84%.

Learn more on Stoichiometry here: brainly.com/question/9372758