JAMB Chemistry 1991 Past Questions & Explanations

\textbf{Reaction:}

\[\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\]

\vspace{2mm}

\textbf{Given:}

\begin{align*}

\text{Volume of Na}_2\text{CO}_3 &= 10 \text{ cm}^3 \\

\text{Concentration of Na}_2\text{CO}_3 &= 0.1 \text{ mol/dm}^3 \\

\text{G.M.V at s.t.p} &= 22400 \text{ cm}^3/\text{mol}

\end{align*}

\vspace{2mm}

\textbf{Step 1: Calculate moles of Na}_2\text{CO}_3

\[n = M \times V = 0.1 \, \text{mol/dm}^3 \times \frac{10}{1000} \, \text{dm}^3\]

\[n = 0.001 \text{ mol}\]

\vspace{2mm}

\textbf{Step 2: Relate moles of Na}_2\text{CO}_3 \text{ to moles of CO}_2

From the balanced equation, mole ratio of $\text{Na}_2\text{CO}_3 : \text{CO}_2 = 1:1$

\[n(\text{CO}_2) = n(\text{Na}_2\text{CO}_3) = 0.001 \text{ mol}\]

\vspace{2mm}

\textbf{Step 3: Calculate volume of CO}_2 \text{ at s.t.p}

\[V(\text{CO}_2) = n \times \text{G.M.V} = 0.001 \text{ mol} \times 22400 \text{ cm}^3/\text{mol}\]

\[\boxed{V(\text{CO}_2) = 22.4 \text{ cm}^3}\]

\vspace{2mm}

\textbf{Answer: Option B (22.4 cm}^3\textbf{)}

\textbf{Solution}

Using Faraday's First Law,

\[Q=It\]

where

\[I=1.5\,\mathrm{A}, \qquad t=4.00\times3600=14400\,\mathrm{s}.\]

Therefore,

\[Q=1.5\times14400=21600\,\mathrm{C}.\]

Using

\[m=\frac{QM}{nF},\]

where

\[m=13.3\,\mathrm{g},\quad M=118.7\,\mathrm{g\,mol^{-1}},\quad F=96500\,\mathrm{C\,mol^{-1}}.\]

Rearranging,

\[n=\frac{QM}{mF}.\]

Substituting the values,

\[n=\frac{21600\times118.7}{13.3\times96500} \approx2.\]

Hence, the oxidation state of tin in the molten salt is

\[\boxed{+2}.\]

A solution has pH greater than 7 if it is alkaline (basic). Salts formed from a strong base and a weak acid undergo hydrolysis to produce an alkaline solution.

\textbf{Solution}

The half-equation is

\[\mathrm{MnO_4^- + 8H^+ + ne^- \rightarrow Mn^{2+} + 4H_2O}\]

Determine the oxidation state of manganese in $\mathrm{MnO_4^-}$.

\[x + 4(-2) = -1\]

\[x - 8 = -1\]

\[x = +7\]

The oxidation state of manganese in $\mathrm{Mn^{2+}}$ is

\[+2.\]

Therefore, the change in oxidation state is

\[+7 \rightarrow +2.\]

The number of electrons gained is

\[7 - 2 = 5.\]

Hence,

\[n = 5.\]

The balanced reduction half-equation is

\[\boxed{\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}}\]

\[\boxed{\text{Correct Answer: D }(5)}\]

\textbf{Solution}

The reaction is

\[\mathrm{2H_2S_{(g)} + SO_2{}_{(g)} \rightarrow 3S_{(s)} + 2H_2O_{(l)}}.\]

Determine the oxidation states of sulphur.

In $\mathrm{H_2S}$,

\[2(+1)+x=0,\]

\[x=-2.\]

In $\mathrm{SO_2}$,

\[x+2(-2)=0,\]

\[x=+4.\]

In elemental sulphur,

\[\mathrm{S},\]

the oxidation state is

\[0.\]

Therefore,

\[\mathrm{H_2S}:\quad -2 \rightarrow 0,\]

which is an increase in oxidation state. Hence, $\mathrm{H_2S}$ is oxidized and acts as the \textbf{reducing agent (reductant)}.

Also,

\[\mathrm{SO_2}:\quad +4 \rightarrow 0,\]

which is a decrease in oxidation state. Hence, $\mathrm{SO_2}$ is reduced and acts as the \textbf{oxidizing agent (oxidant)}.

Therefore, the reaction is a redox reaction in which

\[\boxed{\mathrm{SO_2}\ \text{is the oxidant and}\ \mathrm{H_2S}\ \text{is the reductant}.}\]

\[\boxed{\text{Correct Answer: B}}\]

As a catalyst, MnO2 lowers activation energy of a reaction.

Q = mct = (50 x 4.18 x 3.4) / (1000)
= 0.71606 kj
1.1 g of call2 contains
(1.1) / (111) = 0.01 moles of call2
If 0.01 liberates 0.7106 RS
1 will liberate (0.7106 x 1) / (0.01)
Endothermic = 71.06 = -71.1

Since ΔH is negative, the reaction is exothermic, so a low temperature favors the forward reaction and increases the yield of products.

Also, the number of moles of gas decreases from 2 moles on the left to 1.5 moles on the right. According to Le Chatelier's Principle, increasing the pressure favors the side with fewer moles of gas, which is the product side.

Therefore, the conditions for maximum conversion are:

• Low temperature and High pressure

A change in pressure affects an equilibrium only when the number of moles of gaseous reactants and products are different.

Nitrous oxide (N₂O) supports combustion and can rekindle a glowing splint, just like oxygen. It decomposes on heating to release oxygen, which helps the splint burn again.

Sodium carbonate (Na₂CO₃) is a Group I (alkali metal) carbonate, which is thermally stable and can be melted without decomposing. Carbonates of most other metals (such as CaCO₃, MgCO₃, and ZnCO₃) decompose on heating to form the metal oxide and carbon(IV) oxide.

When potassium trioxonitrate(V) (KNO₃) is heated, it decomposes to produce oxygen gas, potassium nitrite, and oxygen.

Oxygen is commonly prepared in the laboratory by heating potassium trioxochlorate(V) in the presence of manganese(IV) oxide, which acts as a catalyst. The catalyst speeds up the decomposition without being consumed.

When carbon(IV) oxide (CO₂) is bubbled into calcium hydroxide (limewater), a white precipitate of calcium trioxocarbonate(IV) (CaCO₃) is formed.

Mercury and silver are less reactive metals, so their trioxonitrate(V) salts decompose on heating to give the free metal, nitrogen dioxide, and oxygen.

Copper(II) tetraoxosulphate(VI) (CuSO₄) is widely used as a fungicide to prevent and control fungal diseases in crops.

But-2-ene exhibits geometric (cis-trans) isomerism because each carbon atom of the double bond is attached to two different groups. This allows the molecule to exist in two different spatial arrangements: cis and trans.

\[\mathrm{C_3H_4 + 2Br_2 \rightarrow C_3Br_4 + 2H_2}\]

\[(3 \times 12) + (4 \times 1) = 40\]

\[(2 \times 80) = 160\]

\[40\,\mathrm{g}\ \text{of}\ \mathrm{C_3H_4}\ \text{reacts with}\ 320\,\mathrm{g}\ \text{of}\ \mathrm{Br_2}.\]

\[10\,\mathrm{g}\ \text{will react with}\]

\[\frac{10}{40} \times 320 = 80\,\mathrm{g}.\]

Ethene reacts with concentrated tetraoxosulphate(VI) acid (H₂SO₄) to form ethyl hydrogen sulphate. On dilution with water and warming, it is hydrolysed to produce ethanol.

Unlike soaps, detergents do not form insoluble scum in hard water. Instead, they form soluble salts with calcium and magnesium ions, allowing them to lather and clean effectively even in hard water.