For a given unbalanced reaction, $MnO_2 + HCl \rightarrow MnCl_2 + H_2O + Cl_2$, which is the limiting reagent if the initial amount for each reactant is $100 \ g$? [Molar masses: $MnO_2 = 86.9 \ g/mol$; $HCl = 36.5 \ g/mol$]

When $400 \ mL$ of $0.2 \ M \ H_2SO_4$ solution is mixed with $600 \ mL$ of $0.1 \ M \ NaOH$ solution,the increase in temperature of the final solution is $...... \times 10^{-2} \ K$. (Round off to the Nearest Integer).
$\left[ \text{Use } : H^{+}_{(aq)} + OH^{-}_{(aq)} \rightarrow H_2O : \Delta_{r}H = -57.1 \ kJ \ mol^{-1} \right]$
Specific heat of $H_2O = 4.18 \ J \ K^{-1} \ g^{-1}$
Density of $H_2O = 1.0 \ g \ cm^{-3}$
Assume no change in volume of solution on mixing.

The set of numerical coefficients that balances the equation $K_2CrO_4 + HCl \to K_2Cr_2O_7 + KCl + H_2O$ is

Combustion of $50 \ mL$ methane is carried out with $150 \ mL$ of air containing $21 \%$ oxygen by volume. What will be the total volume of the gaseous mixture after the reaction? (Assume water is in the gaseous state and air contains $79 \%$ nitrogen by volume).
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(g)}$

For the given reaction:
$CaCO_{3} + 2HCl \to CaCl_{2} + H_{2}O + CO_{2}$
If $90 \ g$ of $CaCO_{3}$ is added to $300 \ mL$ of $HCl$ solution which contains $38.55\%$ $HCl$ by mass and has a density of $1.13 \ g \ mL^{-1}$, then which of the following options is correct?
[Given molar masses of $H$, $Cl$, $Ca$, and $O$ are $1$, $35.5$, $40$, and $16 \ g \ mol^{-1}$ respectively]

When $20 \ g$ of $CaCO_3$ is treated with $20 \ g$ of $HCl$,the mass of $CO_2$ formed would be: (in $g$)