The maximum number of moles of $CaSO_4$ produced by the reaction of $0.5 \ mol$ of $H_2SO_4$ with $0.2 \ mol$ of $Ca(OH)_2$ is ........

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

Sulphuric acid reacts with sodium hydroxide as follows:
$H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$
What will be the amount of sodium sulphate formed, when $1 \ L$ of $0.2 \ M$ sulphuric acid is allowed to react with $1 \ L$ of $0.2 \ M$ sodium hydroxide solution (in $g$)?

One mole of $Cl_{2(g)}$ was passed into $2 \text{ L}$ of cold $2 \text{ M}$ $KOH$ solution. After the reaction, the concentrations of $Cl^{-}$, $ClO^{-}$ and $OH^{-}$ are respectively (assume volume remains constant):

On reaction of $38.1 \ g$ iodine and $28.4 \ g$ chlorine,a mixture of $ICl$ and $ICl_3$ is obtained. What will be the total number of moles after the reaction?

At $NTP$,if $100 \ mL$ of $N_2$ and $100 \ mL$ of $H_2$ are mixed together in a container,what will be the volume of $NH_3$ formed in $mL$?