2 $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
$\begin{array}{cc}\left(\text { a) } \quad \underset{\mathrm{MnO}_4^{-}}{ }+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \longrightarrow\right. & 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O} \\ \downarrow \\ \text { Oxidation }=+7 & +2\end{array}$ State In this reaction manganese is getting reduced. So, this reaction is not a disproportionation reaction (b.) $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
[JEE Main 2019
Some Basic Concepts of Chemistry
228708
For a reaction, identify di-hydrogen $\left(\mathrm{H}_2\right)$ as a limiting reagent in the following reaction mixture.
1 $56 \mathrm{~g}$ of $\mathrm{N}_2+10 \mathrm{~g}$ of $\mathrm{H}_2$
2 $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
3 $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$
4 $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ of $\mathrm{H}_2$
Explanation:
$\begin{array}{ccc} \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow & 2 \mathrm{NH}_3(\mathrm{~g}) \\ 2 \times 14 \mathrm{~g} & 3 \times 2 \mathrm{~g} & 2(14+3) \\ 28 \mathrm{~g} & 6 \mathrm{~g} & 34 \mathrm{~g} \end{array}$ $28 \mathrm{~g} \mathrm{~N}_2$ requires $6 \mathrm{~g} \mathrm{H}_2 gas.$ 56 \mathrm{~g} \text { of } \mathrm{N}_2 \text { requires } \frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 56 \mathrm{~g}=12 \mathrm{~g} \text { of } \mathrm{H}_2 $12 \mathrm{~g}$ of $\mathrm{H}_2$ gas is required for $56 \mathrm{~g}$ of $\mathrm{N}_2$ gas but only $10 \mathrm{~g}$ of $\mathrm{H}_2$ gas is present in option (a). Hence, $\mathrm{H}_2$ gas is the limiting reagent (b.) $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
[JEE Main 2019
Some Basic Concepts of Chemistry
228711
Volume of water needed to mix with $10 \mathrm{~mL}$ $10 \mathrm{~N} \mathrm{HNO} \mathrm{H}_3$ to get $0.1 \mathrm{~N} \mathrm{HNO} \mathrm{HO}_3$ is
1 $1000 \mathrm{~mL}$
2 $990 \mathrm{~mL}$
3 $1010 \mathrm{~mL}$
4 $10 \mathrm{~mL}$
Explanation:
Given, $\mathrm{N}_1=10 \mathrm{~N}, \mathrm{~V}_1=\mathrm{mL}, \mathrm{N}_2=0.1 \mathrm{~N}$ and $\mathrm{V}_2=$ ? By law of conservation $\begin{aligned} & \mathrm{N}_1 \mathrm{~V}_1=\mathrm{N}_2 \mathrm{~V}_2 \\ & 10 \times 10=0.1(10+\mathrm{V}) \\ & \mathrm{V}=\frac{10 \times 10}{0.1}-10=1000-10=990 \mathrm{~mL} \end{aligned}$
AIIMS-2017
Some Basic Concepts of Chemistry
228713
On combustion of $x-g$ of ethanol in bomb calorimeter, y-joules of heat energy is produced. The heat of combustion of ethanol $\left(\Delta \mathbf{H}_{\text {comb }}\right)$ is-
2 $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
$\begin{array}{cc}\left(\text { a) } \quad \underset{\mathrm{MnO}_4^{-}}{ }+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \longrightarrow\right. & 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O} \\ \downarrow \\ \text { Oxidation }=+7 & +2\end{array}$ State In this reaction manganese is getting reduced. So, this reaction is not a disproportionation reaction (b.) $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
[JEE Main 2019
Some Basic Concepts of Chemistry
228708
For a reaction, identify di-hydrogen $\left(\mathrm{H}_2\right)$ as a limiting reagent in the following reaction mixture.
1 $56 \mathrm{~g}$ of $\mathrm{N}_2+10 \mathrm{~g}$ of $\mathrm{H}_2$
2 $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
3 $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$
4 $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ of $\mathrm{H}_2$
Explanation:
$\begin{array}{ccc} \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow & 2 \mathrm{NH}_3(\mathrm{~g}) \\ 2 \times 14 \mathrm{~g} & 3 \times 2 \mathrm{~g} & 2(14+3) \\ 28 \mathrm{~g} & 6 \mathrm{~g} & 34 \mathrm{~g} \end{array}$ $28 \mathrm{~g} \mathrm{~N}_2$ requires $6 \mathrm{~g} \mathrm{H}_2 gas.$ 56 \mathrm{~g} \text { of } \mathrm{N}_2 \text { requires } \frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 56 \mathrm{~g}=12 \mathrm{~g} \text { of } \mathrm{H}_2 $12 \mathrm{~g}$ of $\mathrm{H}_2$ gas is required for $56 \mathrm{~g}$ of $\mathrm{N}_2$ gas but only $10 \mathrm{~g}$ of $\mathrm{H}_2$ gas is present in option (a). Hence, $\mathrm{H}_2$ gas is the limiting reagent (b.) $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
[JEE Main 2019
Some Basic Concepts of Chemistry
228711
Volume of water needed to mix with $10 \mathrm{~mL}$ $10 \mathrm{~N} \mathrm{HNO} \mathrm{H}_3$ to get $0.1 \mathrm{~N} \mathrm{HNO} \mathrm{HO}_3$ is
1 $1000 \mathrm{~mL}$
2 $990 \mathrm{~mL}$
3 $1010 \mathrm{~mL}$
4 $10 \mathrm{~mL}$
Explanation:
Given, $\mathrm{N}_1=10 \mathrm{~N}, \mathrm{~V}_1=\mathrm{mL}, \mathrm{N}_2=0.1 \mathrm{~N}$ and $\mathrm{V}_2=$ ? By law of conservation $\begin{aligned} & \mathrm{N}_1 \mathrm{~V}_1=\mathrm{N}_2 \mathrm{~V}_2 \\ & 10 \times 10=0.1(10+\mathrm{V}) \\ & \mathrm{V}=\frac{10 \times 10}{0.1}-10=1000-10=990 \mathrm{~mL} \end{aligned}$
AIIMS-2017
Some Basic Concepts of Chemistry
228713
On combustion of $x-g$ of ethanol in bomb calorimeter, y-joules of heat energy is produced. The heat of combustion of ethanol $\left(\Delta \mathbf{H}_{\text {comb }}\right)$ is-
2 $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
$\begin{array}{cc}\left(\text { a) } \quad \underset{\mathrm{MnO}_4^{-}}{ }+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \longrightarrow\right. & 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O} \\ \downarrow \\ \text { Oxidation }=+7 & +2\end{array}$ State In this reaction manganese is getting reduced. So, this reaction is not a disproportionation reaction (b.) $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
[JEE Main 2019
Some Basic Concepts of Chemistry
228708
For a reaction, identify di-hydrogen $\left(\mathrm{H}_2\right)$ as a limiting reagent in the following reaction mixture.
1 $56 \mathrm{~g}$ of $\mathrm{N}_2+10 \mathrm{~g}$ of $\mathrm{H}_2$
2 $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
3 $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$
4 $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ of $\mathrm{H}_2$
Explanation:
$\begin{array}{ccc} \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow & 2 \mathrm{NH}_3(\mathrm{~g}) \\ 2 \times 14 \mathrm{~g} & 3 \times 2 \mathrm{~g} & 2(14+3) \\ 28 \mathrm{~g} & 6 \mathrm{~g} & 34 \mathrm{~g} \end{array}$ $28 \mathrm{~g} \mathrm{~N}_2$ requires $6 \mathrm{~g} \mathrm{H}_2 gas.$ 56 \mathrm{~g} \text { of } \mathrm{N}_2 \text { requires } \frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 56 \mathrm{~g}=12 \mathrm{~g} \text { of } \mathrm{H}_2 $12 \mathrm{~g}$ of $\mathrm{H}_2$ gas is required for $56 \mathrm{~g}$ of $\mathrm{N}_2$ gas but only $10 \mathrm{~g}$ of $\mathrm{H}_2$ gas is present in option (a). Hence, $\mathrm{H}_2$ gas is the limiting reagent (b.) $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
[JEE Main 2019
Some Basic Concepts of Chemistry
228711
Volume of water needed to mix with $10 \mathrm{~mL}$ $10 \mathrm{~N} \mathrm{HNO} \mathrm{H}_3$ to get $0.1 \mathrm{~N} \mathrm{HNO} \mathrm{HO}_3$ is
1 $1000 \mathrm{~mL}$
2 $990 \mathrm{~mL}$
3 $1010 \mathrm{~mL}$
4 $10 \mathrm{~mL}$
Explanation:
Given, $\mathrm{N}_1=10 \mathrm{~N}, \mathrm{~V}_1=\mathrm{mL}, \mathrm{N}_2=0.1 \mathrm{~N}$ and $\mathrm{V}_2=$ ? By law of conservation $\begin{aligned} & \mathrm{N}_1 \mathrm{~V}_1=\mathrm{N}_2 \mathrm{~V}_2 \\ & 10 \times 10=0.1(10+\mathrm{V}) \\ & \mathrm{V}=\frac{10 \times 10}{0.1}-10=1000-10=990 \mathrm{~mL} \end{aligned}$
AIIMS-2017
Some Basic Concepts of Chemistry
228713
On combustion of $x-g$ of ethanol in bomb calorimeter, y-joules of heat energy is produced. The heat of combustion of ethanol $\left(\Delta \mathbf{H}_{\text {comb }}\right)$ is-
2 $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
$\begin{array}{cc}\left(\text { a) } \quad \underset{\mathrm{MnO}_4^{-}}{ }+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \longrightarrow\right. & 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O} \\ \downarrow \\ \text { Oxidation }=+7 & +2\end{array}$ State In this reaction manganese is getting reduced. So, this reaction is not a disproportionation reaction (b.) $2 \mathrm{NaBr}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_2$ Oxidation $+1 \quad+1$ state This is just an example of a displacement reaction not a disproportionation reaction. $\begin{array}{lcc}\text { (c) } 2 \mathrm{KMnO}_4 & \longrightarrow & \mathrm{K}_2 \mathrm{MnO}_4+ \\ \text { Oxidation }+7 & \mathrm{MnO}_2+\mathrm{O}_2\end{array}$ state This indicate manganese is only getting reduced. So, this reaction is not a disproportionation reaction. $\begin{array}{llc}(\mathrm{d}) \quad 2 \mathrm{CuBr} \longrightarrow & \mathrm{CuBr}_2+\mathrm{Cu} \\ \text { Oxidation }+1 & +2 & 0\end{array}$ state In this reaction copper is getting both oxidized as well as reduced Therefore, this reaction is an example of a dioproportionation reaction.
[JEE Main 2019
Some Basic Concepts of Chemistry
228708
For a reaction, identify di-hydrogen $\left(\mathrm{H}_2\right)$ as a limiting reagent in the following reaction mixture.
1 $56 \mathrm{~g}$ of $\mathrm{N}_2+10 \mathrm{~g}$ of $\mathrm{H}_2$
2 $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
3 $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$
4 $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ of $\mathrm{H}_2$
Explanation:
$\begin{array}{ccc} \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow & 2 \mathrm{NH}_3(\mathrm{~g}) \\ 2 \times 14 \mathrm{~g} & 3 \times 2 \mathrm{~g} & 2(14+3) \\ 28 \mathrm{~g} & 6 \mathrm{~g} & 34 \mathrm{~g} \end{array}$ $28 \mathrm{~g} \mathrm{~N}_2$ requires $6 \mathrm{~g} \mathrm{H}_2 gas.$ 56 \mathrm{~g} \text { of } \mathrm{N}_2 \text { requires } \frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 56 \mathrm{~g}=12 \mathrm{~g} \text { of } \mathrm{H}_2 $12 \mathrm{~g}$ of $\mathrm{H}_2$ gas is required for $56 \mathrm{~g}$ of $\mathrm{N}_2$ gas but only $10 \mathrm{~g}$ of $\mathrm{H}_2$ gas is present in option (a). Hence, $\mathrm{H}_2$ gas is the limiting reagent (b.) $35 \mathrm{~g}$ of $\mathrm{N}_2+8 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g} \mathrm{~N}_2$ required $6 \mathrm{~g}$ of $\mathrm{H}_2$ $35 \mathrm{~g} \mathrm{~N}_2$ required $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 35 \mathrm{gH}_2$ $\Rightarrow \quad 7.5 \mathrm{~g}$ of $\mathrm{H}_2$ Here, $\mathrm{H}_2$ gas does not act as limiting reagent since $7.5 \mathrm{~g}$ of $\mathrm{H}_2$ is present in reaction mixture similarly, in option (c) and (d), $\mathrm{H}_2$ doesnot act as limiting reagent. For $14 \mathrm{~g}$ of $\mathrm{N}_2+4 \mathrm{~g}$ of $\mathrm{H}_2$. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts will $6 \mathrm{~g}$ of $\mathrm{H}_2$. $14 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $\frac{6}{28} \times 14 \mathrm{~g}$ of $\mathrm{H}_2 \Rightarrow 3 \mathrm{~g}$ of $\mathrm{H}_2$ For $28 \mathrm{~g}$ of $\mathrm{N}_2+6 \mathrm{~g}$ or $\mathrm{H}_2$, i.e. $28 \mathrm{~g}$ of $\mathrm{N}_2$ reacts with $6 \mathrm{~g}$ of $\mathrm{H}_2$.
[JEE Main 2019
Some Basic Concepts of Chemistry
228711
Volume of water needed to mix with $10 \mathrm{~mL}$ $10 \mathrm{~N} \mathrm{HNO} \mathrm{H}_3$ to get $0.1 \mathrm{~N} \mathrm{HNO} \mathrm{HO}_3$ is
1 $1000 \mathrm{~mL}$
2 $990 \mathrm{~mL}$
3 $1010 \mathrm{~mL}$
4 $10 \mathrm{~mL}$
Explanation:
Given, $\mathrm{N}_1=10 \mathrm{~N}, \mathrm{~V}_1=\mathrm{mL}, \mathrm{N}_2=0.1 \mathrm{~N}$ and $\mathrm{V}_2=$ ? By law of conservation $\begin{aligned} & \mathrm{N}_1 \mathrm{~V}_1=\mathrm{N}_2 \mathrm{~V}_2 \\ & 10 \times 10=0.1(10+\mathrm{V}) \\ & \mathrm{V}=\frac{10 \times 10}{0.1}-10=1000-10=990 \mathrm{~mL} \end{aligned}$
AIIMS-2017
Some Basic Concepts of Chemistry
228713
On combustion of $x-g$ of ethanol in bomb calorimeter, y-joules of heat energy is produced. The heat of combustion of ethanol $\left(\Delta \mathbf{H}_{\text {comb }}\right)$ is-
