What volume (ml) of 3.0 m naoh is required to react with 0.8024-g copper (ii) nitrate? what mass of copper (ii) hydroxide will form, assuming 100% yield?

The reaction between NaOH and Cu (NO₃) ₂ is as follows

2NaOH + Cu (NO₃) ₂ - - - > 2NaNO₃ + Cu (OH) ₂

Q1)

stoichiometry of NaOH to Cu (NO₃) ₂ is 2:1

this means that 2 mol of NaOH reacts with 1 mol of Cu (NO₃) ₂

the mass of Cu (NO₃) ₂ reacted - 0.8024 g

molar mass of Cu (NO₃) ₂ is 187.56 g/mol

therefore the number of Cu (NO₃) ₂ moles that have reacted

- 0.8024 g / 187.56 g/mol = 0.00427 mol

according to the stoichiometry, number of NaOH moles - 0.00427 mol x 2

then number of NaOH moles that have reacted - 0.00855 mol

In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution

Then volume required for 0.00855 mol - 1000 x 0.00855 / 3 = 2.85 mL

2.85 mL of 3.0 M NaOH is required for this reaction

Q2)

Assuming that there's 100 % yield of Cu (OH) ₂, we can directly calculate the mass of Cu (OH) ₂ formed from the number of moles of reactants that were used up.

Stoichiometry of Cu (NO₃) ₂ to Cu (OH) ₂ is 1:1

this means that 1 mol of Cu (NO₃) ₂ gives a yield of 1 mol of Cu (OH) ₂

the number of Cu (NO₃) ₂ moles that reacted - 0.00427 mol

Therefore an equal amount of moles of Cu (OH) ₂ were formed

Then amount of Cu (OH) ₂ moles produced - 0.00427 mol

Mass of Cu (OH) ₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g

A mass of 0.42 g of Cu (OH) ₂ was formed in this reaction