In a certain operation, $358 \mathrm{g}$ of ${\mathrm{TiCl}}_4$ is reacted with $96 \mathrm{g}$ of $\mathrm{Mg}$. Calculate $\%$of yield of $\mathrm{Ti}$ if $32 \mathrm{g}$ of $\mathrm{Ti}$ is actually obtained [At. wt. $\mathrm{Ti} =48, \mathrm{Mg}=24$]

The minimum amount of ${\mathrm{Ag}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)$ required to produce sufficient oxygen for the complete combustion of ${\mathrm{C}}_2{\mathrm{H}}_2$ which produces $11.2 \mathrm{L}$ of ${\mathrm{CO}}_2$ at $\mathrm{STP}$ after combustion is:

$[\mathrm{Ag}=108]$

${\mathrm{Ag}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)\to 2\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)+ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{O}}_2\left(\mathrm{g}\right)$

${\mathrm{C}}_2{\mathrm{H}}_2+\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

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?

$29.5 \mathrm{mg}$ of an organic compound containing nitrogen was digested according to Kjeldahl method and the evolved ammonia was absorbed in $20 \mathrm{mL}$ of  $0.1 \mathrm{M} \mathrm{HCl}$ solution. The excess of acid required $15 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ solution for complete neutralisation. The percentage of nitrogen in the compound is:

A solution containing $2.675 \mathrm{g}$ of ${\mathrm{CoCl}}_3.6{\mathrm{NH}}_3 \left(\mathrm{molar} \mathrm{mass} =267.5\mathrm{g} {\mathrm{mol}}^{-1}\right)$ is passed through a cation exchanger. The chloride ions obtained in the solution were treated with an excess of ${\mathrm{AgNO}}_3$ to give $4.31 \mathrm{g}$ of $\mathrm{AgCl}$ $(\mathrm{molar} \mathrm{mass}=143.5 \mathrm{g} {\mathrm{mol}}^{-1})$. The formula of the complex is: $\left(\mathrm{Ag}=108\mathrm{u}\right)$

$\text{ Hint:molesof }{\mathrm{Cl}}^{-}=\mathrm{moles}\text{ of }\mathrm{AgCl}=\frac{4.31}{143.5}=0.03.$

$\because 0.01\mathrm{mol} \mathrm{of} {\mathrm{CoCl}}_3.6{\mathrm{NH}}_3 \mathrm{produces} 0.03 \mathrm{mol} \mathrm{of} {\mathrm{Cl}}^{-}$

$\therefore 1\text{ mole of }{\mathrm{CoCl}}_3·6{\mathrm{NH}}_3\text{ contains }3{\mathrm{Cl}}^{-}\text{anions }$