Help With parts D and E please.

This problem refers to part $B$ and $C$ of the procedure

a$.$ How many moles of NaOH are needed to neutralize the excess nitric acid from Part A$?$

excess $=0\mathrm{.}4553352214$mal$+N{O}_3$

$0\mathrm{.}46$ mal $HN{O}_3\frac{1\text{m}\text{o}\text{l}\text{N}\text{a}\text{O}\text{H}}{\text{i}\text{m}\text{a}\text{l}\text{H}\text{i}\text{N}\text{e}\text{z}}=0\mathrm{.}46$molNaOH

$.HN{O}_{3}aq+-$NaOHaq$NON{O}_{3}aq+H_{2}O(1)$

b$.$ How many moles of NaOH are needed to convert your copper $($II$)$ nitrate to copper

$($II$)$ hydroxide in Reaction $2?$

$0\mathrm{.}46$molNaOH$\frac{1\text{m}\text{a}\text{l}\text{C}\text{u}(N{O}_3{)}_2}{2\text{m}\text{a}\text{l}\text{N}\text{a}\text{O}\text{H}}=0\mathrm{.}227667611$molCu$(N{e}_3{)}_2$

$=0\mathrm{.}23$molCu$(N{O}_3{)}_2$

$2$NaOH:$1C(N{O}_3{)}_2$

v$.m=\frac{\text{m}\text{o}\text{l}}{v}*v$

$\frac{M}{(mol)}=\frac{\text{m}\text{o}\text{l}*V}{l}$

c$.$ The NaOH solution contains $6$ moles per liter $(6\mathrm{.}0$MNaOH$).$ What volume $($in $mL)$ is needed to provide the total quantity calculated of NaOH needed in part $B?$

$M=6\mathrm{.}0$mollL,$V=\frac{M}{(mol)},V=26\mathrm{.}35420487L$

$V=m,V=26L$

mal$=0\mathrm{.}23$mol,$V=\frac{6\mathrm{.}0\text{m}\text{o}\text{l}\text{l}}{0\mathrm{.}23\text{m}\text{a}\text{l}},V=0\mathrm{.}026$mLNaOt

$V=0\mathrm{.}026$mLNaOHI

d$.$ Suppose you add just enough NaOH so that all the $Cu(N{O}_3{)}_2$ is converted to solid $Cu(OH{)}_2$ by Reaction $2$ and all the remaining $HN{O}_3$ is neutralized in Reaction $3.$ How many moles of $Cu(OH{)}_2$ is produced?

$Cu(N{O}_3{)}_2+2$NaOH$Cu(OH{)}_2+2NaN{O}_3$

e$.$ If all of the $Cu(OH{)}_2$ is converted to CuO, how many moles of CuO are formed?