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 $1 - 0.4 = 0.6 \ dm^{3}$.