Common salt obtained from seawater contains $96\%$ $\mathrm{NaCl}$ by mass. The approximate number of molecules of $\mathrm{NaCl}$ present in $10.0 \mathrm{g}$ of the common salt is: (Atomic weight, $\mathrm{Na}=23$)

$2\mathrm{KI}+{\mathrm{I}}_2+22{\mathrm{HNO}}_3\to 2{\mathrm{HIO}}_3+2{\mathrm{KIO}}_3+22{\mathrm{NO}}_2+10{\mathrm{H}}_2\mathrm{O}$

If $3$ moles of $\mathrm{KI}$ & $2$ moles ${\mathrm{I}}_2$ are reacted with an excess of ${\mathrm{HNO}}_3$, the volume of ${\mathrm{NO}}_2$ gas evolved at $\mathrm{NTP}$ is:

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}$

Consider the following statements:

1. If all the reactants are not taken in their stoichiometric ratio, then at least one of the reactant will be left behind.
2. $2$ moles of ${\mathrm{H}}_2\left(\mathrm{g}\right)$ and $3$ moles of ${\mathrm{O}}_2\left(\mathrm{g}\right)$ produce $2$ moles of water.
3. Equal wt. of carbon and oxygen are taken to produce ${\mathrm{CO}}_2$, then ${\mathrm{O}}_2$ is the limiting reagent.

The above statements $1, 2, 3$ respectively are:

$(\mathrm{T}=\text{ True, }\mathrm{F}=\text{ False })$