$KMnO_4$ 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?

$A$ closed $10 \, L$ vessel contains $1 \, L$ water gas $(1:1 \, CO : H_2)$ and $9 \, L$ air ($20 \% \, O_2$ by volume) at $STP$. The contents of the vessel are ignited. The number of moles of $CO_2$ in the vessel is closest to $.... \, mol$.

When $22.4 \ L$ of $H_{2(g)}$ is mixed with $11.2 \ L$ of $Cl_{2(g)}$,each at $S.T.P.$,the moles of $HCl_{(g)}$ formed is equal to

Calcium carbonate reacts with aqueous $HCl$ to give $CaCl_2$ and $CO_2$ according to the reaction given below: $CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
What mass of $CaCl_2$ will be formed when $250 \ mL$ of $0.76 \ M$ $HCl$ reacts with $1000 \ g$ of $CaCO_3$? Name the limiting reagent. Calculate the number of moles of $CaCl_2$ formed in the reaction.

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

If $30 \ mL$ of $H_2$ and $20 \ mL$ of $O_2$ react to form water,what is left at the end of the reaction?