What volume (ml) of 3.0m 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 mole of Cu(NO₃)₂ gives a yield of  1 mole of Cu(OH)₂

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

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.

did anyone actually do this? like if anyone has submitted this i rly need .