The mass of $N_2F_4$ produced by the reaction of $2.0 \ mol$ of $NH_3$ and $8.0 \ mol$ of $F_2$ is $0.5 \ mol.$ What is the per cent yield?
$2NH_3 + 5F_2 \to N_2F_4 + 6HF$

$1.0 \ g$ of $Mg$ is burnt with $0.28 \ g$ of $O_2$ in a closed vessel. Which reactant is left in excess and how much?

$CaCO_{3(s)} + 2 HCl_{(aq)} \rightarrow CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
Consider the above reaction,what mass of $CaCl_2$ will be formed if $250 \ mL$ of $0.76 \ M \ HCl$ reacts with $1000 \ g$ of $CaCO_3$ (in $g$)?
(Given: Molar mass of $Ca, C, O, H$ and $Cl$ are $40, 12, 16, 1$ and $35.5 \ g \ mol^{-1}$,respectively)

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).
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(g)}$

$A + 2B + 3C \rightleftharpoons AB_2C_3$. Reaction of $6.0 \ g$ of $A$,$6.0 \times 10^{23}$ atoms of $B$,and $0.036 \ mol$ of $C$ yields $4.8 \ g$ of compound $AB_2C_3$. If the atomic mass of $A$ and $C$ are $60$ and $80 \ amu$,respectively,the atomic mass of $B$ is .............. $amu$ (Avogadro no. $= 6 \times 10^{23}$)

$KMnO_4$ 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?