How many grams of calcium chloride are required to produce $14.35 \ g$ of $AgCl$? (Atomic weights: $Ca = 40, Ag = 108, Cl = 35.5$)

The ratio of mass percentage (w/w) of $C : H$ in a hydrocarbon is $12 : 1$. It has two carbon atoms. The weight (in $g$) of $CO_2(g)$ formed when $3.38 \ g$ of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in $g \ mol^{-1}$, $C : 12, H : 1, O : 16$)

An unknown chlorohydrocarbon has $3.55\%$ of chlorine. If each molecule of the hydrocarbon has one chlorine atom only,the number of chlorine atoms present in $1\,g$ of chlorohydrocarbon is (Atomic wt. of $Cl = 35.5\,u$; Avogadro constant $= 6.023 \times 10^{23}\,mol^{-1}$)

The yield of each step as actually carried out in the laboratory is given in the reaction scheme below. Calculate the overall percentage yield of the reaction.
$Me_3C-CO-CH_3$ $\xrightarrow{58\% (A)} Me_3C-CO-CH_2Br$ $\xrightarrow{54\% (B)} Me_3C-CH(OH)-CH_2Br$ $\xrightarrow{68\% (C)} Me_3C-CH(O)CH_2$

What is meant by stoichiometry?

Match the following items in List-$I$ with the corresponding results in List-$II$ (at $STP$):

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$