The heat of combustion of ethylene $C_2H_{4(g)}$ is $-1420 \, kJ/mol$. The number of litres of $C_2H_4$ at $NTP$ that would evolve $355 \, kJ$ on combustion is:

$0.01 \, \text{mole}$ of $H_3PO_x$ is completely neutralized by $0.56 \, \text{g}$ of $KOH$. Hence:

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

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

$X \ g$ of $Ca$ [atomic mass $= 40$] dissolves completely in concentrated $HCl$ solution to produce $5.04 \ L$ of $H_2$ gas at $STP$. The value of $X$ is closest to

$A$ $6 \ L$ sample of hard water has a total hardness of $200 \ ppm$ of $Mg^{2+}$ in the form of bicarbonates and chlorides. This sample is treated with $Na_2CO_3$ to eliminate all temporary hardness,and the new hardness is found to be $\frac{1000}{3} \ ppm$ of $CaCO_3$. Find the mass of $MgCl_2$ in the $6 \ L$ sample in $mg$. $(d_{H_2O} = 1 \ g/mL)$

Aqueous $HCl$ reacts with $MnO_{2(s)}$ to form $MnCl_{2(aq)}$, $Cl_{2(g)}$ and $H_{2}O_{(l)}$. What is the weight (in $g$) of $Cl_{2}$ liberated when $8.7 \ g$ of $MnO_{2(s)}$ is reacted with excess aqueous $HCl$ solution? (Given Molar mass in $g \ mol^{-1}$: $Mn=55$, $Cl=35.5$, $O=16$, $H=1$)