Chemical stoichiometry Questions in English

(B) The neutralization reaction is $HCl + NaOH \longrightarrow NaCl + H_2O$.
Number of milliequivalents of $HCl = x \times y$.
Number of milliequivalents of $NaOH = y \times x$.
Since the milliequivalents are equal, the solution is neutral and the number of milliequivalents of $NaCl$ formed is $xy$.
Total volume of solution = $x \ mL$ $(HCl)$ + $y \ mL$ $(NaOH)$ + $(x+y) \ mL$ (distilled water) = $2(x+y) \ mL$.
$\text{Normality} = \frac{\text{Total milliequivalents of salt}}{\text{Total volume in } mL} = \frac{xy}{2(x+y)} \ N$.

(A) $1$. Calculate the molar mass of $NaOH$: $Na(23) + O(16) + H(1) = 40 \ g/mol$.
$2$. Calculate the number of moles of $NaOH$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.
$3$. Convert the volume of the solution from $mL$ to $L$: $V = \frac{250 \ mL}{1000} = 0.25 \ L$.
$4$. Calculate the molarity $(M)$: $M = \frac{n}{V} = \frac{0.1 \ mol}{0.25 \ L} = 0.4 \ M$.

(B) The balanced chemical equation for the titration is:
$Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_2O_{(\ell)}$
Calculate the millimoles of $Ba(OH)_2$ used:
$n_{Ba(OH)_2} = M \times V_{(mL)} = 0.1 \ M \times 25.0 \ mL = 2.5 \ mmol$
From the stoichiometry, $1 \ mol$ of $Ba(OH)_2$ reacts with $2 \ mol$ of $HCl$:
$n_{HCl} = 2 \times n_{Ba(OH)_2} = 2 \times 2.5 \ mmol = 5.0 \ mmol$
Calculate the mass of $HCl$ in $mg$:
$Mass = n \times Molar \ mass = 5.0 \ mmol \times 36.5 \ mg/mmol = 182.5 \ mg$
Given $x \ mg = 182.5 \ mg$, we need to express this as $x \times 10^{-1}$:
$182.5 = 1825 \times 10^{-1}$
Therefore, $x = 1825$.

(A) The balanced chemical equation for the reaction is: $MnO_{2(s)} + 4HCl_{(aq)} \rightarrow MnCl_{2(aq)} + Cl_{2(g)} + 2H_{2}O_{(l)}$
Calculate the molar mass of $MnO_{2}$: $55 + (2 \times 16) = 87 \ g \ mol^{-1}$.
Calculate the number of moles of $MnO_{2}$ used: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{8.7 \ g}{87 \ g \ mol^{-1}} = 0.1 \ mol$.
According to the stoichiometry of the reaction, $1 \ mol$ of $MnO_{2}$ produces $1 \ mol$ of $Cl_{2}$.
Therefore, $0.1 \ mol$ of $MnO_{2}$ will produce $0.1 \ mol$ of $Cl_{2}$.
Calculate the molar mass of $Cl_{2}$: $2 \times 35.5 = 71 \ g \ mol^{-1}$.
Calculate the weight of $Cl_{2}$ liberated: $0.1 \ mol \times 71 \ g \ mol^{-1} = 7.1 \ g$.

(B) The balanced chemical equation is: $2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_{2(g)}$.
From the stoichiometry, $2 \ \text{moles of } Al$ produce $3 \ \text{moles of } H_{2(g)}$.
Therefore, $1 \ \text{mole of } Al$ produces $\frac{3}{2} = 1.5 \ \text{moles of } H_{2(g)}$.
At $STP$, $1 \ \text{mole of any gas occupies } 22.4 \ L$.
So, $1.5 \ \text{moles of } H_{2(g)} = 1.5 \times 22.4 \ L = 33.6 \ L \ H_{2(g)}$.
Thus, $33.6 \ L \ H_{2(g)}$ at $STP$ is produced for every mole of $Al$ that reacts.

(B) $1$. Determine the empirical formula: The ratio of mass percentage of $C : H$ is $12 : 1$. Dividing by atomic masses $(C=12, H=1)$, the mole ratio is $(12/12) : (1/1) = 1 : 1$. The empirical formula is $CH$.
$2$. Determine the molecular formula: The hydrocarbon has two carbon atoms, so the molecular formula is $(CH)_2 = C_2H_2$ (Ethyne).
$3$. Write the balanced combustion reaction: $2C_2H_2 + 5O_2 \to 4CO_2 + 2H_2O$.
$4$. Calculate moles of hydrocarbon: Molar mass of $C_2H_2 = (2 \times 12) + (2 \times 1) = 26 \ g/mol$. Moles in $3.38 \ g = 3.38 / 26 = 0.13 \ mol$.
$5$. Calculate moles of $CO_2$: From the balanced equation, $1 \ mol$ of $C_2H_2$ produces $2 \ mol$ of $CO_2$. Therefore, $0.13 \ mol$ of $C_2H_2$ produces $0.13 \times 2 = 0.26 \ mol$ of $CO_2$.
$6$. Calculate mass of $CO_2$: Mass $= 0.26 \ mol \times 44 \ g/mol = 11.44 \ g$.

(B) The thermal decomposition reaction of silver carbonate is: $Ag_2CO_3(s) \to 2Ag(s) + CO_2(g) + 1/2 O_2(g)$.
Since $CO_2$ and $O_2$ are gases, the solid residue obtained is metallic silver $(Ag)$.
Molar mass of $Ag_2CO_3 = (2 \times 108) + 12 + (3 \times 16) = 216 + 12 + 48 = 276 \text{ g mol}^{-1}$.
Number of moles of $Ag_2CO_3 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{2.76 \text{ g}}{276 \text{ g mol}^{-1}} = 0.01 \text{ mol}$.
According to the stoichiometry of the reaction, $1 \text{ mole}$ of $Ag_2CO_3$ produces $2 \text{ moles}$ of $Ag$.
Therefore, moles of $Ag$ produced $= 2 \times 0.01 \text{ mol} = 0.02 \text{ mol}$.
Mass of $Ag$ residue $= \text{moles} \times \text{molar mass} = 0.02 \text{ mol} \times 108 \text{ g mol}^{-1} = 2.16 \text{ g}$.

(D) The balanced chemical equation for the reaction of iron with steam is: $3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$.
From the stoichiometry of the reaction, $4$ moles of $H_2O$ react with $3$ moles of $Fe$.
The molar mass of $H_2O = (2 \times 1) + 16 = 18\text{ g mol}^{-1}$.
Thus, $4 \times 18\text{ g} = 72\text{ g}$ of steam reacts with $3 \times 56\text{ g} = 168\text{ g}$ of iron.
Using the unitary method, $18\text{ g}$ of steam will react with: $\frac{168\text{ g Fe}}{72\text{ g H}_2\text{O}} \times 18\text{ g H}_2\text{O} = \frac{168}{4} = 42\text{ g}$ of iron.

(D) The chemical reaction is: $CO_{2}(g) + C(s) \rightarrow 2CO(g)$.
Let the initial volume of $CO_{2}$ be $1 \ dm^{3}$.
Let $x$ be the volume of $CO_{2}$ that reacts with carbon.
According to the stoichiometry of the reaction, $1 \ mole$ of $CO_{2}$ produces $2 \ moles$ of $CO$. Thus, $x \ dm^{3}$ of $CO_{2}$ will produce $2x \ dm^{3}$ of $CO$.
The remaining volume of $CO_{2}$ is $(1 - x) \ dm^{3}$.
The total volume of the gaseous mixture is the sum of the remaining $CO_{2}$ and the produced $CO$: $(1 - x) + 2x = 1 + x$.
Given that the final volume is $1.4 \ dm^{3}$, we have $1 + x = 1.4$, which gives $x = 0.4 \ dm^{3}$.
Therefore, the volume of $CO$ produced is $2x = 2(0.4) = 0.8 \ dm^{3}$.
The volume of remaining $CO_{2}$ is $1 - 0.4 = 0.6 \ dm^{3}$.

(A) The chemical equation for the complete combustion of $n$-butane $(C_4H_{10})$ is:
$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$
First, calculate the molar mass of $n$-butane $(C_4H_{10})$:
$M(C_4H_{10}) = 4 \times 12 + 10 \times 1 = 48 + 10 = 58 \ g/mol$
Calculate the number of moles of $n$-butane in $116 \ g$:
$n = \frac{116 \ g}{58 \ g/mol} = 2 \ mol$
From the balanced equation, $2 \ mol$ of $C_4H_{10}$ produces $8 \ mol$ of $CO_2$.
Calculate the mass of $CO_2$ produced:
Molar mass of $CO_2 = 12 + 2 \times 16 = 44 \ g/mol$
Mass of $CO_2 = 8 \ mol \times 44 \ g/mol = 352 \ g$.

(C) The balanced chemical equation for the synthesis of ammonia is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
From the equation, $3 \text{ moles}$ of $H_2$ produce $2 \text{ moles}$ of $NH_3$.
Molar mass of $NH_3 = 14 + 3 \times 1 = 17 \text{ g/mol}$.
Moles of $NH_3$ produced $= \frac{3.4 \text{ g}}{17 \text{ g/mol}} = 0.2 \text{ mol}$.
Since $2 \text{ moles}$ of $NH_3$ require $3 \text{ moles}$ of $H_2$, $0.2 \text{ moles}$ of $NH_3$ require $\frac{3}{2} \times 0.2 = 0.3 \text{ moles}$ of $H_2$.
Number of $H_2$ molecules $= 0.3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{23}$.
Note: The provided options appear incorrect based on the calculation. The closest value is $1.8066 \times 10^{23}$.

(D) The formula for percentage atom economy is:
$\text{Atom Economy} = \left( \frac{\text{Formula weight of desired product}}{\text{Sum of formula weights of all reactants}} \right) \times 100\%$
Given:
Formula weight of desired product = $65 \text{ u}$
Sum of formula weights of all reactants = $130 \text{ u}$
Calculation:
$\text{Atom Economy} = \left( \frac{65}{130} \right) \times 100\%$
$\text{Atom Economy} = 0.5 \times 100\% = 50\%$
Therefore, the correct option is $D$.

(B) Step $1$: Calculate the molar mass of $CO_2$. Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \text{ g/mol}$.
Step $2$: Calculate the number of moles of $CO_2$. $n = \frac{\text{mass}}{\text{molar mass}} = \frac{99 \text{ g}}{44 \text{ g/mol}} = 2.25 \text{ mol}$.
Step $3$: Calculate the volume at $STP$. At $STP$, $1 \text{ mole}$ of any gas occupies $22.4 \text{ L}$.
Volume $= n \times 22.4 \text{ L/mol} = 2.25 \times 22.4 \text{ L} = 50.4 \text{ L}$.

(A) Step $1$: Identify the given values: Molarity $(M) = 0.1 \text{ mol/L}$, Volume $(V) = 100 \text{ ml} = 0.1 \text{ L}$, Molar mass $(MW) = 106 \text{ g/mol}$.
Step $2$: Use the formula for Molarity: $M = \frac{\text{mass}}{MW \times V(\text{in L})}$.
Step $3$: Rearrange to find mass: $\text{mass} = M \times MW \times V$.
Step $4$: Substitute the values: $\text{mass} = 0.1 \text{ mol/L} \times 106 \text{ g/mol} \times 0.1 \text{ L} = 1.06 \text{ g}$.

(B) Step $1$: Write the balanced chemical equation for the thermal decomposition of potassium chlorate:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
Step $2$: From the stoichiometry, $2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
Step $3$: At $STP$, $1 \text{ mole}$ of any gas occupies $22.4 \text{ L}$. Therefore, $11.2 \text{ L}$ of $O_2$ corresponds to $n = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles}$ of $O_2$.
Step $4$: Using the mole ratio, $3 \text{ moles}$ of $O_2$ are produced by $2 \text{ moles}$ of $KClO_3$. So, $0.5 \text{ moles}$ of $O_2$ are produced by $\frac{2}{3} \times 0.5 = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \text{ moles}$ of $KClO_3$.

(D) The balanced chemical equation for the thermal decomposition of potassium chlorate is:
$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$
From the stoichiometry of the reaction, $2 \text{ moles}$ of $KClO_3$ produce $3 \text{ moles}$ of $O_2$.
At $STP$, $1 \text{ mole}$ of any ideal gas occupies $22.4 \text{ L}$.
Therefore, $22.4 \text{ L}$ of $O_2$ corresponds to $1 \text{ mole}$ of $O_2$.
Using the mole ratio: $\frac{2 \text{ moles } KClO_3}{3 \text{ moles } O_2} = \frac{x \text{ moles } KClO_3}{1 \text{ mole } O_2}$.
Solving for $x$: $x = \frac{2}{3} \text{ moles}$ of $KClO_3$.

(A) The formula for atom economy is defined as:
$\text{Atom Economy} = \left( \frac{\text{Formula weight of desired product}}{\text{Sum of formula weights of all reactants}} \right) \times 100\%$
Given:
$\text{Atom Economy} = 50\% = 0.5$
$\text{Sum of formula weights of all reactants} = 274 \text{ u}$
Let the formula weight of the desired product be $x$.
$0.5 = \frac{x}{274 \text{ u}}$
$x = 0.5 \times 274 \text{ u} = 137 \text{ u}$
Thus, the formula weight of the desired product is $137 \text{ u}$.

(B) The partial pressure of a gas is given by $P_i = x_i P_{total}$, where $x_i$ is the mole fraction.
For the partial pressures to be equal $(P_{Ne} = P_{CH_4})$, their mole fractions must be equal $(x_{Ne} = x_{CH_4})$.
Since $x_i = \frac{n_i}{n_{total}}$, equal mole fractions imply equal number of moles $(n_{Ne} = n_{CH_4})$.
Let the number of moles be $n$. The mass of $Ne$ is $m_{Ne} = n \times M_{Ne} = n \times 20 \ \text{g/mol}$.
The mass of $CH_4$ is $m_{CH_4} = n \times M_{CH_4} = n \times 16 \ \text{g/mol}$.
The ratio by mass is $\frac{m_{Ne}}{m_{CH_4}} = \frac{20n}{16n} = \frac{5}{4}$ or $5:4$.

(B) Step $1$: Identify the symbols and valencies of the elements. Tin $(IV)$ has a valency of $+4$ $(Sn^{4+})$ and Oxide has a valency of $-2$ $(O^{2-})$.
Step $2$: Use the criss-cross method to determine the formula. The subscript of $Sn$ becomes $2$ and the subscript of $O$ becomes $4$, resulting in $Sn_2O_4$.
Step $3$: Simplify the ratio of the subscripts to the lowest whole number. Dividing both by $2$ gives $SnO_2$.