Chemical Equations and Stoichiometry

In the Haber process $30 \mathrm{litre}$ of dihydrogen and $30 \mathrm{litres}$ of dinitrogen were taken for reaction which yielded only  $50\%$ of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?

In Haber's process, $30 \mathrm{litres}$ of dihydrogen and $30 \mathrm{litres}$ of dinitrogen were taken for the reaction, which yielded only $50\mathrm{\%}$ of the expected product. What will be the composition of the gaseous mixture under the aforesaid condition in the end?

Assuming full decomposition, the volume of  ${\mathrm{CO}}_2$ released at STP on heating  $9.85\hspace{0.17em}\mathrm{g}\hspace{0.17em}\mathrm{of}\hspace{0.17em}{\mathrm{BaCO}}_3$ (Atomic mass, $Ba=137$) will be:

Liquid benzene  $\left({\mathrm{C}}_6{\mathrm{H}}_6\right)$ burns in oxygen according to the equation

$2{\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+15{\mathrm{O}}_2\left(\mathrm{g}\right)\to 12{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

How many litres of  ${\mathrm{O}}_2$ at STP are needed to complete the combustion of $39 \mathrm{g}$ of liquid benzene?  $(\mathrm{Molecular}\mathrm{weight}\mathrm{of}{\mathrm{O}}_2=32,{\mathrm{C}}_6{\mathrm{H}}_6=78)$

Liquid benzene $\left({\mathrm{C}}_6{\mathrm{H}}_6\right)$ burns in oxygen according to the equation,

$2{\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right)+15{\mathrm{O}}_2\left(\mathrm{g}\right)\to 12{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
How many litres of ${\mathrm{O}}_2$ at $\mathrm{STP}$ are needed to complete the combustion of $39 \mathrm{g}$ of liquid benzene? (Molecular weight of ${\text{O}}_2=32\mathrm{,} {\mathrm{C}}_6{\mathrm{H}}_6=78$)

The volume of hydrogen liberated at STP by treating $2.4 \mathrm{g}$ of magnesium with excess of hydrochloric acid is ......$\times {10}^{-2} \mathrm{L}$. Given Molar volume of gas is $22.4 \mathrm{L}$ at STP. Molar mass of magnesium is $24 \mathrm{g} {\mathrm{mol}}^{-1}$
 

$1 \mathrm{g}$ of a carbonate $\left({\mathrm{M}}_2{\mathrm{CO}}_3\right)$ on treatment with excess $\mathrm{HCl}$ produces $0.01 \mathrm{mol}$ of ${\mathrm{CO}}_2$. The molar mass of ${\mathrm{M}}_2{\mathrm{CO}}_3$ is ${\mathrm{gmol}}^{-1}$. (Nearest integer)

The volume (in $\mathrm{mL}$) of $0.1 \mathrm{M} {\mathrm{AgNO}}_3$ required for complete precipitation of chloride ions present in $20 \mathrm{mL}$ of $0.01 \mathrm{M}$ solution of $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2$ as silver chloride is_______

One mole of ${\mathrm{P}}_4$ reacts with $8$ moles of ${\mathrm{SOCl}}_2$ to give $4$ moles of $\mathrm{A}$, $\mathrm{x}$ mole of ${\mathrm{SO}}_2$ and $2$ moles of $\mathrm{B}$. $\mathrm{A}, \mathrm{B}$ and $\mathrm{x}$ respectively are

If $5$ moles of ${\mathrm{BaCl}}_2$ is mixed with $2$ moles of ${\mathrm{Na}}_3{\mathrm{PO}}_4,$ the maximum number of moles of ${\mathrm{Ba}}_3{\left({\mathrm{PO}}_4\right)}_2$formed is $\_\_\_\_\_\_\_\_\_$ (Nearest integer)

Consider the following sequence of reactions and the percentage yield is mentioned against them

$$\begin{gathered} \mathrm{Ca} + 2\mathrm{C}\stackrel{}{\to }{\mathrm{CaC}}_2 80\% \\ {\mathrm{CaC}}_2+{\mathrm{N}}_2\stackrel{}{\to }{\mathrm{CaCN}}_2+\mathrm{C} 50\% \\ {\mathrm{CaCN}}_2+3{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }2{\mathrm{NH}}_3+{\mathrm{CaCO}}_3 75\% \end{gathered}$$

If $36 \mathrm{g}$ of carbon is allowed reacts with excess of $\mathrm{Ca}$ in the first reaction, then the mass of calcium carbonate formed is [Assume${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{N}}_2$ are in excess amount ]

Consider the reaction  $4{\mathrm{HNO}}_3\left(\mathrm{l}\right) + 3\mathrm{KCl}\left(\mathrm{s}\right) \to {\mathrm{Cl}}_2\left(\mathrm{g}\right) + \mathrm{NOCl}\left(\mathrm{g}\right) + 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) + 3{\mathrm{KNO}}_3\left(\mathrm{s}\right)$

The amount of ${\mathrm{HNO}}_3$ required to produce $110.0 \mathrm{g} \mathrm{of} {\mathrm{KNO}}_3$ is : 

(Given : Atomic masses of $\mathrm{H}, \mathrm{O}, \mathrm{N} \mathrm{and} \mathrm{K} \mathrm{are} 1, 16, 14 \mathrm{and} 39,$ respectively.) 

In the given reaction

$\mathrm{X}+\mathrm{Y}+3\mathrm{Z}\stackrel{}{\rightleftharpoons }{\mathrm{XYZ}}_3$

If one mole of each $\mathrm{X}$and $\mathrm{Y}$ with $0.05 \mathrm{mol}$ of $\mathrm{Z}$ gives compound ${\mathrm{XYZ}}_3$.

(Given: Atomic masses of $\mathrm{X}, \mathrm{Y},$and $\mathrm{Z}$ are $10, 20 \mathrm{and} 30 \mathrm{amu}$ respectively)

The yield of ${\mathrm{XYZ}}_3$ is $-------\mathrm{g}.$

Round off to the nearest integer.

Balance the following equation and choose the quantity which is the sum of the coefficients of the products:

$.....{\mathrm{CS}}_2+ ......{\mathrm{Cl}}_2\stackrel{}{\to }......{\mathrm{CCl}}_4+.....{\mathrm{S}}_2{\mathrm{Cl}}_2$

$1 \mathrm{g}$ sample of chalk is completely neutralised by $100 \mathrm{mL}$ of $0.04\mathrm{N} \mathrm{HCl}$ solution. $\%\mathrm{w}/\mathrm{w} {\mathrm{CaCO}}_3$ in the sample is 

$10\mathrm{g} \mathrm{of} {\mathrm{CaCO}}_3$ is treated with $250 \mathrm{ml}$ of $2\mathrm{M} \mathrm{HCl}$ solution. What is the normality of $\mathrm{HCl}$ in resulting solution?

Chlorine can be prepared by reacting $\mathrm{HCl}$ with ${\mathrm{MnO}}_2$​. The reaction is represented by the equation,
${\mathrm{MnO}}_2\left(\mathrm{s}\right)+4\mathrm{HCl}(\mathrm{aq}.)\to {\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathrm{MnCl}}_2(\mathrm{aq}.)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
 Assuming that the reaction goes to completion, what mass of $\mathrm{conc}. \mathrm{HCl}$ solution ($36\% \mathrm{HCl}$ by mass) is needed to produce $2.5 \mathrm{g} {\mathrm{Cl}}_2$?