Calculate the amount of ${\mathrm{H}}_2$ which is left unreacted in the given reaction:

$2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$

If $8\mathrm{gm}$ of ${\mathrm{H}}_2$ is mixed with $16\mathrm{gm}$ of ${\mathrm{O}}_2$.

Production of iron in blast furnace follows the following equation

${\mathrm{Fe}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)+4\mathrm{CO}\left(\mathrm{g}\right)\to 3\mathrm{Fe}\left(\mathrm{l}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)$

when $4.640\mathrm{kg}$ of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ and $2.520\mathrm{kg}$ of $\mathrm{CO}$ are allowed to react then the amount of iron (in $\mathrm{g}$) produced is :

[Given: Molar Atomic mass $\left({\mathrm{gmol}}^{-1}\right):\mathrm{Fe}=56$

Molar Atomic mass $\left({\mathrm{gmol}}^{-1}\right):\mathrm{O}=16$

Molar Atomic mass $\left({\mathrm{gmol}}^{-1}\right):\mathrm{C}=12$]

In the following reaction:

$\mathrm{A}+2\mathrm{B}+3\mathrm{C}\rightleftharpoons \mathrm{A}{\mathrm{B}}_2{\mathrm{C}}_3$
$6.0 \mathrm{g}$ of $\mathrm{A}$, $6.0\times {10}^{23}\mathrm{ }$ atoms of $\mathrm{B}$ and $0.036 \mathrm{mol}$ of $\mathrm{C}$ reacted and formed $4.8 \mathrm{g}$ of compound $\mathrm{A}{\mathrm{B}}_2{\mathrm{C}}_3$. If the atomic mass of $\mathrm{A}$  and $\mathrm{C}$ are $60$ and $80 \mathrm{amu}$, respectively. What is the atomic mass of $\mathrm{B}$ in $\mathrm{amu}$? (Avogadro number$=6\times {10}^{23}$)

$$\begin{gathered} {\mathrm{N}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{NH}}_{3\left(\mathrm{g}\right)} \\ 20 \mathrm{g} 5 \mathrm{g} \end{gathered}$$

Consider the above reaction, the limiting reagent of the reaction and number of moles of ${\mathrm{NH}}_3$ formed respectively are

$20 \mathrm{ml}$ of acetic acid reacts with $20 \mathrm{ml}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{g}/\mathrm{ml}$ and $0.7 \mathrm{g}/\mathrm{ml}$ respectively. The limiting reagent in this reaction is

$2 \mathrm{mol}$ ${\mathrm{N}}_2$ and $3 \mathrm{mol} {\mathrm{H}}_2$ are allowed to react in a $20 \mathrm{L}$ flask at $400 \mathrm{K}$ and after the complete conversion of ${\mathrm{H}}_2$ to ${\mathrm{NH}}_3$, $10 \mathrm{L} {\mathrm{H}}_2\mathrm{O}$  was added and the temperature is reduced to $300 \mathrm{K}$. The pressure of gas after the reaction is ${\text{N}}_2+3{\text{H}}_2\to 2{\text{NH}}_3$ (assume that all the ${\mathrm{NH}}_3$ formed gets dissolved in water)