$\mathrm{Li}$ metal is one of the few substances that react directly with molecular nitrogen. The balanced equation for the reaction is:

$6\mathrm{Li}\left(\mathrm{s}\right)+{\mathrm{N}}_2\left(\mathrm{g}\right)⟶2{\mathrm{Li}}_3\mathrm{N}\left(\mathrm{s}\right)$

How many grams of the product, lithium nitride, can be prepared from $3.5 \mathrm{g}$ of lithium metal and $8.4 \mathrm{g}$ of molecular nitrogen$?$

Potassium superoxide, ${\mathrm{KO}}_2$, is used in rebreathing gas masks to generate ${\mathrm{O}}_2$. If a reaction vessel contains $0.15$ mol ${\mathrm{KO}}_2$ and $0.10$ mol ${\mathrm{H}}_2\mathrm{O}$, What is the limiting reactant$?$ How many moles of oxygen can be produced?

$2{\mathrm{KO}}_2+2{\mathrm{H}}_2\mathrm{O}⟶2\mathrm{KOH}+{\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{O}}_2$

A 1 g sample of ${\mathrm{KClO}}_3$ was heated under such conditions that a part of it decomposed according to the equation.

(i) $2{\mathrm{KClO}}_3⟶2\mathrm{KCl}+3{\mathrm{O}}_2$

and the remaining underwent change according to the equation

(ii) $4{\mathrm{KClO}}_3⟶3{\mathrm{KClO}}_4+\mathrm{KCl}$

If the amount of ${\mathrm{O}}_2$ evolved was $146.8 \mathrm{mL}$ at $\mathrm{NTP}$, calculate the percentage by weight of ${\mathrm{KClO}}_4$ in the residue.

$64 \mathrm{g}$ of a mixture of $\mathrm{NaCl}$ and $\mathrm{KCl}$ were treated with concentrated sulphuric acid. The total mass of metal sulphates obtained was found to be $76 \mathrm{g}$. What is the mass percents of $\mathrm{NaCl}$ in the mixture. The reactions are,

$2\mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{Na}}_2{\mathrm{SO}}_4+2\mathrm{HCl} ; 2\mathrm{KCl}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{K}}_2{\mathrm{SO}}_4+2\mathrm{HCl}$

$5 \mathrm{g}$ 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