Exercise 1

$\mathrm{Xg}$ of $\mathrm{Ca}$ [atomic mass $=40$] dissolves completely in concentrated $\mathrm{HCl}$ solution to produce $5.40 \mathrm{L}$ of ${\mathrm{H}}_2$ gas at $\mathrm{STP}$. The value of $\mathrm{X}$ is the closest to

What weight of ${\mathrm{CaCO}}_3$ must be decomposed to produce the sufficient quantity of carbon dioxide to convert $21.2 \mathrm{kg}$ of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ completely into ${\mathrm{NaHCO}}_3$. $\mathbf{[}$Atomic mass $\left.\mathrm{Na}=23, \mathrm{Ca}=40\right]$
${\mathrm{CaCO}}_3⟶\mathrm{CaO}+{\mathrm{CO}}_2$

${\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{NaHCO}}_3$

$12 \mathrm{g}$ of alkaline earth metal gives $14.8 \mathrm{g}$ of its nitride. Atomic weight of metal is -

Calculate the amount of $\mathrm{Ni}$ needed in the Mond's process given below.
$\mathrm{Ni}+4\mathrm{CO}⟶\mathrm{Ni}(\mathrm{CO}{)}_4$
If, $\mathrm{CO}$ used in this process is obtained through a process, in which $6 \mathrm{g}$ of carbon is mixed with $44 \mathrm{g} {\mathrm{CO}}_2$.

Equal weight of ${}^{\text{'}}\mathrm{X}^{\text{'}}$ (Atomic weight $=36$) and ${}^{\text{'}}\mathrm{Y}^{\text{'}}$ (Atomic weight $=24$) are reacted to form the compound ${\mathrm{X}}_2{\mathrm{Y}}_3$. Then:

$25.4 \mathrm{g}$ of iodine and $14.2 \mathrm{g}$ of chlorine are made to react completely to yield a mixture of $\mathrm{ICl}$ and ${\mathrm{ICl}}_3$. Calculate the number of moles of $\mathrm{ICl}$and ${\mathrm{ICl}}_3$ formed.

What weights of ${\mathrm{P}}_4{\mathrm{O}}_6$ and ${\mathrm{P}}_4{\mathrm{O}}_{10}$ will be produced by the combustion of $31 \mathrm{g}$ of ${\mathrm{P}}_4$ in $32 \mathrm{g}$ of oxygen leaving no ${\mathrm{P}}_4$ and ${\mathrm{O}}_2$.

The volume of water that must be added to a mixture of $250 \mathrm{ml}$ of $0.6 \mathrm{M} \mathrm{HCl}$ and $750 \mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCl}$ to obtain $0.25 \mathrm{M}$ solution of $\mathrm{HCl}$ is: