What is the maximum amount of $Cl_2$ gas produced when $1 \, g$ of $HCl$ and $1 \, g$ of $MnO_2$ are heated together (in $, g$)?

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below: $C_4H_{10(g)} + \frac{13}{2} O_{2(g)} \rightarrow 4 CO_{2(g)} + 5 H_2O_{(l)}$. If $174.0 \ kg$ of butane is mixed with $320.0 \ kg$ of $O_2$,the volume of water formed in litres is $...........$ (Nearest integer). [$Given$: $(a)$ Molar mass of $C, H, O$ are $12, 1, 16 \ g \ mol^{-1}$ respectively,$(b)$ Density of water $= 1 \ g \ mL^{-1}$]

What will be the volume of the mixture after the reaction? $litre$
$NH_3(4 \ litre) + HCl(1.5 \ litre) \to NH_4Cl(s)$

The maximum amount of $BaSO_4$ precipitated on mixing equal volumes of $BaCl_2$ $(0.5 \ M)$ with $H_2SO_4$ $(1 \ M)$ will correspond to ................. $M$.

$20 \ mL$ of $0.1 \ M$ $HCl$ is added to $30 \ mL$ of $0.1 \ M$ $NaOH$. To this solution,an extra $50 \ mL$ of water was added. What is the molarity of the final solution formed (in $M$)?

$10 \, mL$ of $10 \, M$ $H_2SO_4$ is mixed with $100 \, mL$ of $1 \, M$ $NaOH$ solution. The resultant solution will be: