Chemical stoichiometry Questions in English

(B) According to Avogadro's Law,at the same temperature and pressure,equal volumes of gases contain an equal number of moles.
Since the volume of the vessel is constant,the number of moles of $O_2$ and $SO_2$ will be the same.
Weight of gas = $\text{Number of moles} \times \text{Molar mass}$.
Since the number of moles is the same,the weight of the gas is directly proportional to its molar mass.
Weight of $O_2$ / Weight of $SO_2$ = $\frac{\text{Molar mass of } O_2}{\text{Molar mass of } SO_2} = \frac{32}{64} = \frac{1}{2}$.
Therefore,the weight of oxygen will be $\frac{1}{2}$ that of $SO_2$.

(C) The chemical equation for the decomposition of $CaCO_3$ is:
$CaCO_3(s) \to CaO(s) + CO_2(g)$
The molar mass of $CaCO_3$ is $100 \ g/mol$.
According to the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.
At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$100 \ g$ of $CaCO_3$ produces $22.4 \ L$ of $CO_2$.
Thus,$1 \ g$ of $CaCO_3$ produces $\frac{22.4}{100} = 0.224 \ L$ of $CO_2$.

(C) The density $(d)$ of a gas at $NTP$ is given by the formula: $d = \frac{\text{Molecular Weight}}{\text{Molar Volume at NTP}}$.
Given,Molecular Weight $= 45$ and Molar Volume at $NTP = 22.4 \ L \ mol^{-1}$.
Therefore,$d = \frac{45}{22.4} \approx 2.01 \ g \ L^{-1}$.
Thus,the correct option is $(C)$.

(A) The molar volume of any ideal gas at $STP$ is $22.4 \ L \ mol^{-1}$.
Since the molar mass of $O_2$ is $32 \ g \ mol^{-1}$,we have:
$22.4 \ L \ O_2$ at $STP = 32 \ g$ of $O_2$.
Therefore,$1 \ L \ O_2$ at $STP = \frac{32}{22.4} \approx 1.43 \ g$ of $O_2$.

(C) balanced chemical equation provides the stoichiometric coefficients,which represent the relative number of moles of reactants and products involved in the reaction. This allows for quantitative calculations based on the law of conservation of mass.

(B) The neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Number of millimoles of $H_2SO_4 = M \times V \text{ (in mL)} = 1 \times 10 = 10 \ mmol$.
Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$,$10 \ mmol$ of $H_2SO_4$ requires $20 \ mmol$ of $NaOH$.
Checking the options:
For option $(B)$: $M \times V = 2 \ M \times 10 \ mL = 20 \ mmol$ of $NaOH$.
Thus,$10 \ mL$ of $2 \ M$ $NaOH$ is required for complete neutralization.

(B) The dilution formula is $M_1V_1 = M_2V_2$.
Here,$M_1 = 6.0 \ M$ (concentration of $HCl$ acid).
$V_1$ is the volume of $HCl$ required.
$M_2 = 0.30 \ M$ (required concentration of $H^+$ ions).
$V_2 = 150 \ mL$ (final volume of the solution).
Substituting the values into the equation: $6.0 \times V_1 = 0.30 \times 150$.
$V_1 = \frac{0.30 \times 150}{6.0} = \frac{45}{6.0} = 7.5 \ mL$.

(B) The combustion reaction of methane is: $CH_4 + 2O_2 \to CO_2 + 2H_2O$
The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g/mol$.
Given that the combustion of $2.0 \ g$ of $CH_4$ releases $25 \ kcal$ of heat.
Therefore,the heat of combustion for $1$ mole $(16 \ g)$ of methane is calculated as:
$\text{Heat of combustion} = \frac{25 \ kcal}{2.0 \ g} \times 16 \ g = 200 \ kcal/mol$.

(C) The balanced chemical equation for the combustion of methane is: $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$.
The molar mass of $CH_4$ is $12 + (4 \times 1) = 16 \ g/mol$.
The number of moles of $CH_4$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \ g}{16 \ g/mol} = 0.5 \ mol$.
According to the stoichiometry of the reaction,$1 \ mol$ of $CH_4$ produces $2 \ mol$ of $H_2O$.
Therefore,$0.5 \ mol$ of $CH_4$ will produce $0.5 \times 2 = 1 \ mol$ of $H_2O$.

(D) The balanced chemical equation for the combustion of $1 \ mole$ of gaseous benzene is:
$C_6H_{6(g)} + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$
The change in the number of moles of gaseous species is calculated as:
$\Delta n = \sum n_{p(g)} - \sum n_{r(g)}$
$\Delta n = (6 + 3) - (1 + 7.5) = 9 - 8.5 = +0.5$ or $+\frac{1}{2}$.

(C) Molar mass of butane $(C_4H_{10})$ $= (4 \times 12) + (10 \times 1) = 58 \ g/mol$.
Number of moles of butane in $11.2 \ kg$ $(11200 \ g)$ $= \frac{11200 \ g}{58 \ g/mol} \approx 193.1 \ mol$.
Total energy released by combustion of $11.2 \ kg$ of butane $= 193.1 \ mol \times 2658 \ kJ/mol \approx 513260 \ kJ$.
Daily energy requirement $= 20000 \ kJ$.
Number of days the cylinder will last $= \frac{513260 \ kJ}{20000 \ kJ/day} \approx 25.66 \ days$.
Rounding to the nearest whole number,the cylinder will last for $26 \ days$.

(B) The neutralization reaction for $HCl$ is: $HCl + NaOH \rightarrow NaCl + H_2O$.
Since $1 \text{ mole}$ of $HCl$ reacts with $1 \text{ mole}$ of $NaOH$,$X = 1$.
The neutralization reaction for $H_2SO_4$ is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Since $1 \text{ mole}$ of $H_2SO_4$ reacts with $2 \text{ moles}$ of $NaOH$,$Y = 2$.
Comparing the values,$X = 1$ and $Y = 2$,we get $X = \frac{1}{2}Y$.

(A) The balanced chemical equation for the combustion of ethylene $(C_2H_4)$ is:
$C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$
Molar mass of $C_2H_4 = (2 \times 12) + (4 \times 1) = 28 \ g/mol$ (or $28 \ kg/kmol$).
Molar mass of $3O_2 = 3 \times (2 \times 16) = 96 \ g/mol$ (or $96 \ kg/kmol$).
From the stoichiometry,$28 \ kg$ of $C_2H_4$ requires $96 \ kg$ of $O_2$ for complete combustion.
Therefore,for $2.8 \ kg$ of $C_2H_4$,the weight of $O_2$ required is:
$= \frac{96 \ kg}{28 \ kg} \times 2.8 \ kg = 9.6 \ kg$.

(B) The combustion reaction is $C_2H_6 + O_2 \to CO_2 + H_2O$.
To balance the carbon atoms,we place a coefficient of $2$ before $CO_2$: $C_2H_6 + O_2 \to 2CO_2 + H_2O$.
To balance the hydrogen atoms,we place a coefficient of $3$ before $H_2O$: $C_2H_6 + O_2 \to 2CO_2 + 3H_2O$.
Finally,balancing oxygen atoms gives $C_2H_6 + \frac{7}{2}O_2 \to 2CO_2 + 3H_2O$,or multiplying by $2$ to get whole numbers: $2C_2H_6 + 7O_2 \to 4CO_2 + 6H_2O$.
The coefficients of $CO_2$ and $H_2O$ are $4$ and $6$ respectively.
The ratio is $4:6$,which simplifies to $2:3$.

(A) The chemical reactions are as follows:
$1.$ Reaction with sulphuric acid: $Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2.$ Reaction with sodium hydroxide: $Zn + 2NaOH \to Na_2ZnO_2 + H_2$
From the stoichiometry of both balanced equations,$1 \text{ mole}$ of $Zn$ produces $1 \text{ mole}$ of $H_2$ gas in both cases.
Therefore,for the same amount of $Zn$,the ratio of the volumes of $H_2$ evolved is $1:1$.

(A) The chemical equation for the formation of ozone from oxygen is: $3 O_2 \rightarrow 2 O_3$.
According to the stoichiometry of the reaction,$3$ volumes of oxygen produce $2$ volumes of ozone.
Therefore,$6$ volumes of oxygen will produce: $(6 \times \frac{2}{3}) = 4$ volumes of ozone.

(A) The balanced chemical equation for the reaction of magnesium nitride with water is:
$Mg_3N_2 + 6H_2O \to 3Mg(OH)_2 + 2NH_3$
From the stoichiometry of the reaction,$1 \text{ mole}$ of $Mg_3N_2$ reacts with $6 \text{ moles}$ of $H_2O$ to produce $3 \text{ moles}$ of $Mg(OH)_2$ and $2 \text{ moles}$ of $NH_3$ (ammonia).
Therefore,the correct option is $A$.

(C) The neutralization reaction is $NaOH + HCl \rightarrow NaCl + H_2O$.
For neutralization,the number of equivalents of $NaOH$ must equal the number of equivalents of $HCl$.
Number of equivalents of $HCl = N \times V(L) = 0.1 \ N \times 0.1 \ L = 0.01 \ \text{equivalents}$.
Since the equivalent weight of $NaOH$ is $40 \ g/eq$,the weight required is $W = \text{equivalents} \times \text{equivalent weight} = 0.01 \times 40 = 0.4 \ g$.

(C) To find the normality of the $H_2SO_4$ solution,we use the law of equivalence: $N_1V_1 = N_2V_2$.
Here,$N_1 = 1/10 \ N$,$V_1 = 15 \ mL$,and $V_2 = 12 \ mL$.
Substituting the values: $(1/10) \times 15 = N_2 \times 12$.
$N_2 = 15 / (10 \times 12) = 15 / 120 = 1/8 \ N$.

(C) Initial milliequivalents of $H_2SO_4 = N_1 \times V_1 = 1 \, N \times 30 \, mL = 30 \, meq$.
Final milliequivalents of $H_2SO_4 = N_2 \times V_2 = 0.2 \, N \times 30 \, mL = 6 \, meq$.
Milliequivalents of $NH_3$ required to neutralize the acid = $30 - 6 = 24 \, meq$.
Since $NH_3$ is a base,$1 \, meq$ of $NH_3$ corresponds to $1 \, mmol$ of $NH_3$.
Therefore,$24 \, mmol$ of $NH_3$ is required.
At $N.T.P.$,$1 \, mole$ of gas occupies $22400 \, mL$.
Volume of $NH_3 = 24 \times 10^{-3} \, mol \times 22400 \, mL/mol = 537.6 \, mL$.

(A) For the neutralization reaction,we use the formula $N_1V_1 = N_2V_2$.
Here,$N_1 = 0.2 \ N$,$V_1 = 15 \ mL$,$V_2 = 30 \ mL$,and $N_2 = x$.
Substituting the values: $0.2 \times 15 = x \times 30$.
$3 = 30x$.
$x = \frac{3}{30} = 0.1 \ N$.
Therefore,the concentration of the acid solution is $0.1 \ N$.

(A) For $H_2SO_4$,the normality $(N)$ is related to molarity $(M)$ by the formula $N = n \times M$,where $n$ is the basicity of the acid. For $H_2SO_4$,$n = 2$.
Therefore,$0.1 \, M$ $H_2SO_4$ = $0.1 \times 2 = 0.2 \, N$ $H_2SO_4$.
Using the law of equivalence: $N_1V_1 = N_2V_2$.
Here,$N_1 = 0.2 \, N$ $(H_2SO_4)$,$N_2 = 0.2 \, N$ $(NaOH)$,and $V_2 = 30 \, mL$.
$0.2 \times V_1 = 0.2 \times 30$.
$V_1 = 30 \, mL$.

(D) The dilution formula is $N_1V_1 = N_2V_2$.
Given: $N_1 = 5 \, N$,$V_1 = 1 \, L$,$V_2 = 10 \, L$.
Substituting the values: $5 \, N \times 1 \, L = N_2 \times 10 \, L$.
$N_2 = \frac{5 \times 1}{10} = 0.5 \, N$.

(C) According to the law of equivalence,$N_1V_1 = N_2V_2$.
Here,$N_1 = 5 \ N$,$V_1 = 10 \ mL$,$N_2 = \frac{1}{10} \ N$,and $V_2 = x \ mL$.
Substituting the values: $5 \times 10 = \frac{1}{10} \times x$.
$x = 5 \times 10 \times 10 = 500 \ mL$.

(D) Step $1$: Calculate the normality of $Na_2CO_3$ solution.
Normality of $HCl$ = $\frac{109.5 \ g/L}{36.5 \ g/eq} = 3 \ N$.
Using $N_1V_1 = N_2V_2$ for $Na_2CO_3$ and $HCl$:
$N_{Na_2CO_3} \times 25 \ mL = 3 \ N \times 32.9 \ mL$.
$N_{Na_2CO_3} = \frac{3 \times 32.9}{25} = 3.948 \ N$.
Step $2$: Calculate the volume of $Na_2CO_3$ solution corresponding to $125 \ g$.
Volume = $\frac{\text{mass}}{\text{density}} = \frac{125 \ g}{1.25 \ g/mL} = 100 \ mL$.
Step $3$: Calculate the volume of $0.84 \ N$ $H_2SO_4$ required to neutralize $100 \ mL$ of $3.948 \ N$ $Na_2CO_3$ solution.
$N_3V_3 = N_4V_4$
$3.948 \ N \times 100 \ mL = 0.84 \ N \times V_{H_2SO_4}$.
$V_{H_2SO_4} = \frac{394.8}{0.84} = 470 \ mL$.

(A) The neutralization reaction is governed by the law of equivalence: $N_1V_1 = N_2V_2$.
For $H_2SO_4$,the molarity is $0.05 \, M$. The n-factor for $H_2SO_4$ is $2$.
Therefore,the normality of $H_2SO_4$ is $N_1 = Molarity \times n\text{-factor} = 0.05 \times 2 = 0.1 \, N$.
Given for $NaOH$: $N_2 = 0.13 \, N$ and $V_2 = 80 \, mL$.
Substituting the values into the equation: $0.1 \times V_1 = 0.13 \times 80$.
$V_1 = \frac{0.13 \times 80}{0.1} = 104 \, mL$.

(A) The formula for molarity is given by: $M = \frac{w \times 1000}{M_m \times V(mL)}$.
Here,$M = 0.1 \, M$,$V = 250 \, mL$,and the molar mass of $NaOH$ $(M_m)$ is $23 + 16 + 1 = 40 \, g/mol$.
Substituting the values: $0.1 = \frac{w \times 1000}{40 \times 250}$.
$0.1 = \frac{w \times 1000}{10000}$.
$0.1 = \frac{w}{10}$.
$w = 0.1 \times 10 = 1 \, g$.

(A) The normality equation is $N_1V_1 = N_2V_2$.
Given: $N_1 = 1 \, N$ (for $NaOH$),$V_1 = 100 \, mL$.
For oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$,the equivalent weight is $63$.
Let the weight of oxalic acid be $w$.
Normality of oxalic acid solution = $\frac{w}{\text{Equivalent weight} \times V(L)} = \frac{w}{63 \times 0.1}$.
Equating the two: $1 \times 100 = \left( \frac{w}{63 \times 0.1} \right) \times 1000$.
$100 = \frac{w}{63 \times 0.1} \times 1000 \Rightarrow 100 = \frac{w}{6.3} \times 10 \Rightarrow 10 = \frac{w}{6.3}$.
$w = 6.3 \, g$.

(D) To calculate the volume required for dilution,we use the dilution formula: $N_1V_1 = N_2V_2$.
Here,$N_1 = 10 \, N$ (concentration of concentrated $HCl$),$N_2 = 1 \, N$ (desired concentration),and $V_2 = 1000 \, mL$ (desired volume).
Substituting the values: $10 \, N \times V_1 = 1 \, N \times 1000 \, mL$.
$V_1 = \frac{1000}{10} = 100 \, mL$.
Thus,$100 \, mL$ of concentrated $HCl$ must be diluted to $1000 \, mL$ to obtain the required solution.

(B) The neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Using the normality equation: $N_1V_1 = N_2V_2$.
For $1 \, M$ $H_2SO_4$,the normality is $N_1 = 1 \times 2 = 2 \, N$ (since $n$-factor = $2$).
For $1 \, M$ $NaOH$,the normality is $N_2 = 1 \times 1 = 1 \, N$ (since $n$-factor = $1$).
Substituting the values: $2 \times V_1 = 1 \times 10$.
$V_1 = \frac{10}{2} = 5 \, mL$.

(C) The number of equivalents of $NaOH$ must be equal to the number of equivalents of oxalic acid for neutralization.
Equivalents of oxalic acid = $Normality \times Volume \text{ (in Liters)} = 0.1 \times 0.1 = 0.01 \, eq$.
Equivalents of $NaOH$ = $\frac{\text{Mass}}{\text{Equivalent weight of } NaOH} = \frac{W}{40}$.
Since equivalents are equal: $\frac{W}{40} = 0.01$.
$W = 0.01 \times 40 = 0.4 \, g$.

(B) The neutralization reaction is $NaOH + HCl \rightarrow NaCl + H_2O$.
For neutralization,the number of equivalents of $NaOH$ must equal the number of equivalents of $HCl$.
Equivalents of $HCl = N \times V_{(L)} = 0.1 \, N \times 1.5 \, L = 0.15 \, \text{equivalents}$.
Since the equivalent weight of $NaOH$ is $40 \, g/eq$,the mass of $NaOH$ required is $W = \text{Equivalents} \times \text{Eq. wt.}$.
$W = 0.15 \times 40 = 6 \, g$.

(A) The precipitation reactions are as follows:
For $AgNO_3$: $2Ag^+ + H_2S \rightarrow Ag_2S(s) + 2H^+$. Here,$2 \ mol$ of $Ag^+$ react with $1 \ mol$ of $H_2S$.
For $CuSO_4$: $Cu^{2+} + H_2S \rightarrow CuS(s) + 2H^+$. Here,$1 \ mol$ of $Cu^{2+}$ reacts with $1 \ mol$ of $H_2S$.
Given $100 \ mL$ of $1 \ M$ $AgNO_3$ contains $0.1 \ mol$ of $Ag^+$,requiring $0.05 \ mol$ of $H_2S$.
Given $100 \ mL$ of $1 \ M$ $CuSO_4$ contains $0.1 \ mol$ of $Cu^{2+}$,requiring $0.1 \ mol$ of $H_2S$.
The ratio of $H_2S$ required is $0.05 : 0.1 = 1:2$.

(C) According to the law of equivalence,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid.
$N_1 V_1 = N_2 V_2$
Given $N_1 = 1 \, N$,$V_1 = 20 \, mL$,and $V_2 = 20 \, mL$ for oxalic acid.
$1 \times 20 = N_2 \times 20 \implies N_2 = 1 \, N$
Normality $(N) = \frac{\text{Strength in } g/L}{\text{Equivalent weight}}$
For oxalic acid crystals $(H_2C_2O_4 \cdot 2H_2O)$,the equivalent weight is $63 \, g/eq$.
$\text{Strength} = N \times \text{Equivalent weight} = 1 \times 63 = 63 \, g/L$.

(C) The molar mass of $Na_2CO_3$ is $106 \ g/mol$.
The equivalent mass of $Na_2CO_3$ is $\frac{106}{2} = 53 \ g/eq$.
Normality $(N)$ is given by the formula: $N = \frac{\text{mass in grams}}{\text{equivalent mass} \times \text{volume in Liters}}$.
Given: mass = $0.53 \ g$,volume = $100 \ mL = 0.1 \ L$.
$N = \frac{0.53}{53 \times 0.1} = \frac{0.53}{5.3} = 0.1 \ N = \frac{N}{10}$.

(D) The relationship between Normality $(N)$ and Molarity $(M)$ is given by: $N = M \times \text{n-factor}$.
For $HCl$,the n-factor is $1$,so $2 \ N - HCl = 2 \ M - HCl$.
For $H_2SO_4$,the n-factor is $2$ (since it is a dibasic acid).
To have the same molar concentration $(2 \ M)$,the Normality of $H_2SO_4$ must be: $N = 2 \ M \times 2 = 4 \ N$.
Therefore,$2 \ N - HCl$ has the same molar concentration as $4 \ N - H_2SO_4$.

(C) The reaction is $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Number of equivalents of $H_2SO_4 = N \times V(L) = 1 \times 0.1 = 0.1 \ eq$.
Number of equivalents of $NaOH = M \times n-factor \times V(L) = 1 \times 1 \times 0.1 = 0.1 \ eq$.
Since the number of equivalents of acid $(0.1 \ eq)$ equals the number of equivalents of base $(0.1 \ eq)$,the solution will be $Neutral$.

(B) The balanced chemical equation for the reaction is: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$.
According to the stoichiometry,$1 \, \text{mole}$ of $Na_2CO_3$ reacts with $2 \, \text{moles}$ of $HCl$.
Using the formula $n_1M_1V_1 = n_2M_2V_2$ (where $n$ is the n-factor or stoichiometric coefficient ratio):
$n_1 = 1$ (for $Na_2CO_3$),$M_1 = 0.25 \, M$,$V_1 = 25 \, mL$.
$n_2 = 2$ (for $HCl$),$M_2 = 0.5 \, M$,$V_2 = ?$.
$1 \times 0.25 \times 25 = 2 \times 0.5 \times V_2$.
$6.25 = 1 \times V_2$.
$V_2 = 25 \, mL$.

(C) For complete neutralization,the number of equivalents of the acid must equal the number of equivalents of the base.
Equivalents of $NaOH = \text{Normality} \times \text{Volume (in L)} = \frac{1}{10} \times \frac{25}{1000} = 0.0025 \ eq$.
Since the acid is dibasic,its equivalent weight is $\frac{M}{2}$,where $M$ is the molecular weight.
Equivalents of acid = $\frac{\text{Mass}}{\text{Equivalent Weight}} = \frac{0.16}{M/2} = \frac{0.32}{M}$.
Equating the two: $\frac{0.32}{M} = 0.0025$.
$M = \frac{0.32}{0.0025} = 128 \ g/mol$.

(D) The condition for a neutral solution upon mixing two solutions is ${N_1}{V_1} = {N_2}{V_2}$.
For $NaOH$: ${N_1} = 1 \, N$,${V_1} = 100 \, mL$.
For $H_2SO_4$: ${N_2} = 10 \, N$,${V_2} = 10 \, mL$.
Calculating the milliequivalents for both:
$NaOH: 1 \times 100 = 100 \, meq$.
$H_2SO_4: 10 \times 10 = 100 \, meq$.
Since the number of milliequivalents of acid and base are equal $(100 = 100)$,the resulting solution is neutral.

(A) The formula for normality $(N)$ is given by: $N = \frac{\text{mass of solute (g)}}{\text{equivalent mass} \times \text{volume of solution (L)}}$.
Caustic soda is $NaOH$. Its molar mass is $40 \, g/mol$. Since $NaOH$ is a monoacidic base,its equivalent mass is equal to its molar mass,which is $40 \, g/eq$.
Given: Mass of $NaOH = 4.0 \, g$,Volume of solution = $100 \, mL = 0.1 \, L$.
Substituting the values: $N = \frac{4.0}{40 \times 0.1} = \frac{4.0}{4.0} = 1 \, N$.
Therefore,the correct option is $A$.

(D) To neutralize a solution,the number of gram equivalents of the acid must be equal to the number of gram equivalents of the base.
The formula used is $N_1V_1 = N_2V_2$.
Given: $N_1 = 1 \ N$,$V_1 = 1 \ L$ for $NaOH$.
We need $N_2V_2 = 1 \ N \times 1 \ L = 1 \ \text{equivalent}$.
Option $(D)$ provides $1 \ L$ of $1 \ N \ H_2SO_4$,which contains $1 \ L \times 1 \ N = 1 \ \text{equivalent}$ of acid.
Therefore,$1 \ L$ of $1 \ N \ H_2SO_4$ is required to neutralize $1 \ L$ of $1 \ N \ NaOH$.

(A) The chemical equation for the decomposition of $BaCO_3$ is:
$BaCO_3(s) \to BaO(s) + CO_2(g)$
Calculate the molar mass of $BaCO_3$:
$M(BaCO_3) = 137 + 12 + (3 \times 16) = 137 + 12 + 48 = 197 \ g/mol$
According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ produces $1 \ mol$ of $CO_2$.
At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$197 \ g$ of $BaCO_3$ produces $22.4 \ L$ of $CO_2$ at $S.T.P.$
For $9.85 \ g$ of $BaCO_3$,the volume of $CO_2$ produced is:
$V = \frac{22.4 \ L}{197 \ g} \times 9.85 \ g$
$V = \frac{22.4 \times 9.85}{197} = 1.12 \ L$
Thus,the correct option is $A$.

(D) The relationship between normality and molarity is given by the formula: $\text{Normality} = \text{Molarity} \times \text{Basicity}$.
For phosphoric acid $(H_3PO_4)$,the basicity is $3$ because it can donate $3 \ H^+$ ions per molecule.
Given that the molarity is $1 \ M$,the normality is calculated as: $1 \ M \times 3 = 3 \ N$.

(D) The normality $(N)$ of the solution is given by the formula: $N = \frac{W \times 1000}{Eq.wt \times V(mL)}$.
Here,$N = 1/10$,$V = 1000 \, mL$,and the equivalent weight $(Eq.wt)$ of $H_2SO_4$ is $\frac{98}{2} = 49$.
Substituting the values: $\frac{1}{10} = \frac{W \times 1000}{49 \times 1000}$.
Solving for $W$: $W = \frac{49}{10} = 4.9 \, g$.

(C) For neutralization,the formula is $N_1V_1 = N_2V_2$.
Here,$N_1$ (normality of $NaOH$) $= 1/5 \, N$,$V_1$ (volume of $NaOH$) $= 10 \, mL$.
$N_2$ (normality of $HCl$) $= 1/20 \, N$,$V_2$ (volume of $HCl$) $= ?$.
Substituting the values: $(1/5) \times 10 = (1/20) \times V_2$.
$2 = V_2 / 20$.
$V_2 = 2 \times 20 = 40 \, mL$.

(B) Using the dilution formula: $M_1V_1 = M_2V_2$
Given: $M_1 = 0.15 \, M$,$V_1 = 25 \, mL$,$M_2 = 0.1 \, M$
Substituting the values: $0.15 \times 25 = 0.1 \times V_2$
$V_2 = \frac{0.15 \times 25}{0.1} = 37.5 \, mL$
Volume of water to be added = $V_2 - V_1 = 37.5 \, mL - 25 \, mL = 12.5 \, mL$

(C) The balanced chemical equation for the reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
According to the stoichiometry,$1 \, \text{mole}$ of $H_2SO_4$ reacts with $2 \, \text{moles}$ of $NaOH$.
Using the formula $M_1 V_1 n_2 = M_2 V_2 n_1$,where $n_1$ and $n_2$ are the stoichiometric coefficients:
$0.1 \times V_1 \times 2 = 0.2 \times 40 \times 1$.
$0.2 \times V_1 = 8$.
$V_1 = \frac{8}{0.2} = 40 \, mL$.
Thus,the correct option is $(c)$.

(D) The chemical reaction is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$.
According to the law of equivalence,the number of equivalents of $HCl$ must equal the number of equivalents of $CaCO_3$.
Number of equivalents of $CaCO_3 = \frac{\text{mass}}{\text{equivalent weight}} = \frac{1.0 \, g}{50 \, g/eq} = 0.02 \, eq$.
For $HCl$,the number of equivalents is given by $N \times V(L)$.
$0.1 \, N \times V(L) = 0.02 \, eq$.
$V(L) = \frac{0.02}{0.1} = 0.2 \, L$.
Converting to $cm^3$: $0.2 \, L \times 1000 \, cm^3/L = 200 \, cm^3$.