$5 \mathrm{g}$ sample of ${\mathrm{CuSO}}_4·5{\mathrm{H}}_2\mathrm{O}$ was dissolved in water. ${\mathrm{BaCl}}_2$ solution was mixed in excess to this solution. The precipitate (${\mathrm{BaSO}}_4$) obtained was washed and dried, it weighed $4.66 \mathrm{g}$. What is the % of ${\mathrm{SO}}_4^{2-}$ by weight in the sample?

(mol.wt. of ${\mathrm{BaSO}}_4$ $=233 \mathrm{grams}/\mathrm{mole}$)

Excess of calcium orthophosphate ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$ is reacted with magnesium to form calcium phosphide $\left({\mathrm{Ca}}_3{\mathrm{P}}_2\right)$ with magnesium oxide. Calcium phosphide on reacting with an excess of water liberate phosphine gas (${\mathrm{PH}}_3$) along with calcium hydroxide. The evolved phosphine $\left({\mathrm{PH}}_3\right)$ was burnt in the air (excess of oxygen ) to yield phosphorus pentoxide $\left({\mathrm{P}}_2{\mathrm{O}}_5\right)$ along with water. How many grams of magnesium metaphosphate would be obtained, if $19.2\mathrm{g}$ of magnesium were used for reducing calcium phosphate: 

( Atomic wt. of $\mathrm{Mg}=24, \mathrm{P}=31$ )

${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+ 8\mathrm{Mg}⟶{\mathrm{Ca}}_3{\mathrm{P}}_2+ 8\mathrm{MgO}$
${\mathrm{Ca}}_3{\mathrm{P}}_2+{\mathrm{H}}_2\mathrm{O}⟶\mathrm{Ca}(\mathrm{OH}{)}_2+{\mathrm{PH}}_3$
${\mathrm{PH}}_3+{\mathrm{O}}_2⟶{\mathrm{P}}_2{\mathrm{O}}_5+{\mathrm{H}}_2\mathrm{O}$
$\mathrm{MgO}+{\mathrm{P}}_2{\mathrm{O}}_5⟶\mathrm{Mg}{\left({\mathrm{PO}}_3\right)}_2$

Magnesium metaphosphate

Weight in grams of oxygen in $32.2 \mathrm{g} {\mathrm{Na}}_2{\mathrm{SO}}_4·10{\mathrm{H}}_2\mathrm{O}$ is:

(molecular weight of ${\mathrm{Na}}_2{\mathrm{SO}}_4.10{\mathrm{H}}_2\mathrm{O}=322 \mathrm{g}/\mathrm{mol}$)

The copper metal reacts with Nitric acid based on the reaction

$3\mathrm{Cu}\left(\mathrm{s}\right)+8{\mathrm{HNO}}_3\left(\mathrm{aq}\right)⟶3\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+2\mathrm{NO}\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

If $18.75 \mathrm{g} \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2$ is eventually obtained, then how many grams of Nitric oxide ($\mathrm{NO}$) would have also formed, according to the given equation?