$0.4 \text{ g}$ of dihydrogen is made to react with $7.4 \text{ g}$ of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at $273 \text{ K}$ and $1 \text{ bar}$ pressure is: (in $\text{ L}$)

When $12 \ L$ of $H_2$ and $11.2 \ L$ of $Cl_2$ are mixed and allowed to react according to the equation $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$,what is the final composition of the mixture?

$100 \ mL$ of $0.2 \ M \ H_2SO_4$ is added to $100 \ mL$ of $0.2 \ M \ NaOH$. The resulting solution will be

Consider the following reaction:
$3 PbCl_2 + 2 (NH_4)_3 PO_4 \rightarrow Pb_3(PO_4)_2 + 6 NH_4 Cl$
If $72 \ mmol$ of $PbCl_2$ is mixed with $50 \ mmol$ of $(NH_4)_3 PO_4$,then the amount of $Pb_3(PO_4)_2$ formed is $......... \ mmol$. (nearest integer)

$100 \ mL$ of $0.05 \ M$ $Cu^{2+}$ aqueous solution is added to $1 \ L$ of $0.1 \ M$ $KI$ solution. The number of moles of $I_2$ and $Cu_2I_2$ formed are respectively:

Calculate the maximum possible heat which can be released if a $4 \ L$ aqueous solution is to be formed by mixing appropriate volumes of $2 \ M$ $NaOH$ and $3 \ M$ $H_2SO_4$ solution. Given $\Delta H_{\text{neutralisation, SA/SB}} = -14 \ kcal/mol$.