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: $MnO_2 = 86.9 \ g/mol$; $HCl = 36.5 \ g/mol$]

$92 \ g$ of $NO_{2(g)}$ and $32 \ g$ of $O_{2(g)}$ are taken in a rigid container. At a constant temperature,they react to produce $N_2O_{5(g)}$. The pressure in the container after the reaction is observed to be $\frac{7}{8} P$,where $P$ is the initial pressure of the gaseous mixture. The percentage yield of the reaction is ....$\%$

$KMnO_4$ reacts with $KI$ in basic medium to form $I_2$ and $MnO_2$. When $250 \ mL$ of $0.1 \ M \ KI$ solution is mixed with $250 \ mL$ of $0.02 \ M \ KMnO_4$ in basic medium, what is the number of moles of $I_2$ formed?

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

Calculate the weight of $FeO$ produced from $6.7 \ g$ of $VO$ and $4.8 \ g$ of $Fe_2O_3$ using the reaction: $2VO + 3Fe_2O_3 \rightarrow 6FeO + V_2O_5$ $(V = 51, Fe = 56)$ (in $g$)

$100 \ mL$ 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: