ESAT Chemistry Mock 2

The sum of the percentage abundances must equal $\displaystyle 100\%$ : $\displaystyle x + 30 + y = 100 \implies x = 70 - y$ The relative atomic mass is given by the weighted average: $\displaystyle A_\text{r} = \dfrac{68x + 70(30) + 72y}{100} = 70.2$ Substitute $\displaystyle x = 70 - y$ into the equation: $\displaystyle 68(70 - y) + 2100 + 72y = 7020$ $\displaystyle 4760 - 68y + 2100 + 72y = 7020$ $\displaystyle 6860 + 4y = 7020$ $\displaystyle 4y = 160$ $\displaystyle y = 40$ Therefore, the percentage abundance of $\displaystyle \ce{^{72}X}$ is $\displaystyle 40\%$ .

Statement 1 is incorrect: Group 2 metals act as reducing agents by losing their two outer valence electrons ( $\displaystyle \ce{M -> M^{2+} + 2e-}$ ). Down Group 2, atomic radius and electron shielding increase, lowering the first and second ionisation energies. Consequently, reactivity and reducing power increase down the group, so calcium is a stronger reducing agent than magnesium.

Statement 2 is correct: Magnesium is in Period 3 (three electron shells), whereas calcium is in Period 4 (four electron shells). The additional quantum shell of electrons and increased shielding down the group make the atomic radius of magnesium smaller than that of calcium.

Statement 3 is correct: Magnesium forms $\displaystyle \ce{Mg^{2+}}$ ions and nitrogen forms $\displaystyle \ce{N^{3-}}$ ions, giving the empirical formula $\displaystyle \ce{Mg3N2}$ for magnesium nitride. $\displaystyle \text{Molar mass of } \ce{Mg3N2} = (3 \times 24) + (2 \times 14) = 72 + 28 = 100\text{ g mol}^{-1}$ $\displaystyle \text{Percentage by mass of } \ce{Mg} = \dfrac{72}{100} \times 100\% = 72\%$ Therefore, only statements 2 and 3 are correct.

To find the empirical formula, first determine the molecular formula by counting all atoms in the structure:

1. Carbon atoms:
- The six-membered ring contains 6 carbon atoms.
- There are 2 ester carbonyl carbons and 2 ester methyl carbons ( $\displaystyle 2 \times 2 = 4$ carbons).
- Total carbon atoms = $\displaystyle 6 + 4 = 10$ .

2. Hydrogen atoms:
- In the ring: the two $\displaystyle \ce{C=C}$ double-bond carbons each have 1 hydrogen ( $\displaystyle 2 \times 1 = 2$ ); the two $\displaystyle \ce{CH2}$ carbons each have 2 hydrogens ( $\displaystyle 2 \times 2 = 4$ ); the two $\displaystyle \ce{CH}$ carbons bonded to the ester groups each have 1 hydrogen ( $\displaystyle 2 \times 1 = 2$ ). Total ring hydrogens = $\displaystyle 2 + 4 + 2 = 8$ .
- In the two methyl ester groups: each $\displaystyle \ce{-CH3}$ has 3 hydrogens ( $\displaystyle 2 \times 3 = 6$ ).
- Total hydrogen atoms = $\displaystyle 8 + 6 = 14$ .

3. Oxygen atoms:
- Each ester group contains 2 oxygen atoms ( $\displaystyle 2 \times 2 = 4$ ).
- Total oxygen atoms = 4.

The molecular formula is $\displaystyle \ce{C10H14O4}$ .

Dividing all subscripts by their greatest common divisor (2) gives the simplest whole-number ratio (the empirical formula): $\displaystyle \ce{C5H7O2}$

The component with the lowest boiling point vaporises first: nitrogen ( $\displaystyle -196\,^\circ\text{C}$ ), then argon ( $\displaystyle -186\,^\circ\text{C}$ ), then oxygen ( $\displaystyle -183\,^\circ\text{C}$ ).

To evaluate each statement:

1. Statement 1 is correct:
- Moles of base $\displaystyle \ce{B} = 0.150\text{ mol dm}^{-3} \times 0.0200\text{ dm}^3 = 0.00300\text{ mol}$ .
- Moles of $\displaystyle \ce{HCl} = 0.200\text{ mol dm}^{-3} \times 0.0300\text{ dm}^3 = 0.00600\text{ mol}$ .
- Ratio of moles of $\displaystyle \ce{HCl}$ to base is $\displaystyle \dfrac{0.00600}{0.00300} = 2 : 1$ .
- Since each $\displaystyle \ce{HCl}$ supplies one $\displaystyle \ce{H+}$ ion, $\displaystyle 1\text{ mol}$ of the base reacts with $\displaystyle 2\text{ mol}$ of $\displaystyle \ce{H+}$ .

2. Statement 2 is incorrect:
- Titration measures the total capacity to neutralise acid, not the degree of dissociation (strength). Weak bases also react completely with strong acids in stoichiometric proportions. Furthermore, the precipitation of copper(II) hydroxide, $\displaystyle \ce{Cu(OH)2(s)}$ , only indicates the presence of $\displaystyle \ce{OH-}$ ions, which occur in solutions of both weak and strong bases. Thus, full dissociation cannot be deduced.

3. Statement 3 is correct:
- $\displaystyle M_\text{r} = \dfrac{\text{mass concentration}}{\text{molar concentration}} = \dfrac{11.1\text{ g dm}^{-3}}{0.150\text{ mol dm}^{-3}} = 74$ .

Therefore, statements 1 and 3 only can be deduced.

Oxygen has oxidation state $\displaystyle -2$ . In $\displaystyle SO_2$ , sulfur plus two oxygens must sum to 0, so sulfur is $\displaystyle +4$ . In $\displaystyle SO_4^{2-}$ , sulfur plus four oxygens must sum to $\displaystyle -2$ , so sulfur is $\displaystyle +6$ . The correct option is B.

A liquid has particles close together but in a disordered arrangement, able to move past one another. Diagram B matches this.

Incomplete combustion can form carbon monoxide. High engine temperatures can produce nitrogen oxides from nitrogen and oxygen in air.

Propanoic acid has the formula $\displaystyle \ce{C2H5COOH}$ (or $\displaystyle \ce{C3H6O2}$ ): $\displaystyle M_r(\text{propanoic acid}) = (3 \times 12.0) + (6 \times 1.0) + (2 \times 16.0) = 36.0 + 6.0 + 32.0 = 74.0$ Water ( $\displaystyle \ce{H2O}$ ) has: $\displaystyle M_r(\ce{H2O}) = (2 \times 1.0) + 16.0 = 18.0$ A diol contains two $\displaystyle -\ce{OH}$ groups, so complete esterification with excess propanoic acid requires $\displaystyle 2$ moles of propanoic acid and produces $\displaystyle 1$ mole of diester and $\displaystyle 2$ moles of water: $\displaystyle \text{diol} + 2\,\ce{C2H5COOH} \rightarrow \text{diester} + 2\,\ce{H2O}$ Applying conservation of relative molecular mass: $\displaystyle M_r(\text{diol}) + 2 \times M_r(\text{propanoic acid}) = M_r(\text{diester}) + 2 \times M_r(\ce{H2O})$ $\displaystyle M_r(\text{diol}) + 2(74.0) = 174.0 + 2(18.0)$ $\displaystyle M_r(\text{diol}) + 148.0 = 174.0 + 36.0 = 210.0$ $\displaystyle M_r(\text{diol}) = 210.0 - 148.0 = 62.0$ (This corresponds to ethane-1,2-diol, $\displaystyle \ce{HOCH2CH2OH}$ , which has $\displaystyle M_r = (2 \times 12) + (6 \times 1) + (2 \times 16) = 62$ ).

First, determine the location of the cellular components after the initial spin at 1,000 g.

* Nuclei pellet at 1,000 g, so they are in the pellet.
* Mitochondria (require 15,000 g) and ribosomes (require 100,000 g) do not pellet at 1,000 g, so they remain in the supernatant.

To isolate mitochondria, we must process the supernatant.

* Spinning the supernatant at 15,000 g is sufficient to pellet the mitochondria but not the ribosomes (which require 100,000 g).
* This separates the mitochondria (in the new pellet) from the ribosomes (remaining in the new supernatant).

Therefore, centrifuging the supernatant at 15,000 g and collecting the pellet yields the purest sample. Spinning at 100,000 g (Option D) would pellet both mitochondria and ribosomes, resulting in contamination.

A lilac flame indicates $\displaystyle \ce{K+}$ ; a cream silver halide precipitate indicates $\displaystyle \ce{Br-}$ , so $\displaystyle \ce{KBr}$ fits both tests.

The alloy has a much greater tensile strength but a smaller extension before breaking, so it is stronger and less ductile.

During the electrolysis of dilute $\displaystyle \ce{H2SO4(aq)}$ with inert electrodes:

- At the cathode (negative electrode), hydrogen ions are reduced: $\displaystyle \ce{2H+(aq) + 2e- -> H2(g)}$ .
- At the anode (positive electrode), water is oxidised: $\displaystyle \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}$ .

The overall reaction is $\displaystyle \ce{2H2O(l) -> 2H2(g) + O2(g)}$ .

For every $\displaystyle 4\text{ mol}$ of electrons transferred, $\displaystyle 2\text{ mol}$ of $\displaystyle \ce{H2(g)}$ and $\displaystyle 1\text{ mol}$ of $\displaystyle \ce{O2(g)}$ are produced. Under identical temperature and pressure, the molar volume of both gases is the same, so the volume of $\displaystyle \ce{H2}$ produced is twice the volume of $\displaystyle \ce{O2}$ produced at any given time.

- Line 1 has a gradient twice that of Line 2 ( $\displaystyle 4.8\text{ cm}^3\text{ min}^{-1}$ vs $\displaystyle 2.4\text{ cm}^3\text{ min}^{-1}$ ), so Line 1 represents $\displaystyle \ce{H2(g)}$ and Line 2 represents $\displaystyle \ce{O2(g)}$ . Hydrogen gas is evolved at the cathode, so statement 1 is correct.

- In the external circuit, electrons flow from the anode (where oxidation produces electrons) to the cathode (where reduction consumes electrons). Thus, electrons flow from the Line 2 electrode (anode) towards the Line 1 electrode (cathode). Statement 2 is incorrect.

- The rate of gas volume production is directly proportional to current ( $\displaystyle I$ ): $\displaystyle \dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{V_{\text{m}}}{zF} I$ . Doubling the current doubles the rate of formation of both gases, doubling the gradient of both lines. Statement 3 is correct.

Therefore, statements 1 and 3 only are correct.

1. Relative atomic mass calculation:
From the mass spectrum:
- Isotope with $\displaystyle m/z = 69$ has an abundance of $\displaystyle 60\%$ .
- Isotope with $\displaystyle m/z = 71$ has an abundance of $\displaystyle 40\%$ . $\displaystyle A_\text{r}(\ce{M}) = \dfrac{69 \times 60 + 71 \times 40}{100} = \dfrac{4140 + 2840}{100} = \dfrac{6980}{100} = 69.8$ 2. Empirical formula of the oxide:
- Molar mass of $\displaystyle \ce{O2} = 2 \times 16.0 = 32.0\text{ g mol}^{-1}$ .
- Moles of $\displaystyle \ce{O2} = \dfrac{2.40\text{ g}}{32.0\text{ g mol}^{-1}} = 0.0750\text{ mol}$ .
- Moles of oxygen atoms $\displaystyle \ce{O} = 2 \times 0.0750\text{ mol} = 0.150\text{ mol}$ .
- Moles of $\displaystyle \ce{M} = 0.100\text{ mol}$ .

Molar ratio of $\displaystyle \ce{M} : \ce{O}$ : $\displaystyle \dfrac{n(\ce{M})}{n(\ce{O})} = \dfrac{0.100}{0.150} = \dfrac{2}{3}$ Therefore, the empirical formula of the oxide is $\displaystyle \ce{M2O3}$ .

Matching row: B.

Statement 1 is correct because bromine adds across the C=C bond. Statement 2 is false because addition polymerisation joins monomers without eliminating a small molecule. Statement 3 is correct because alcohols react with sodium to produce hydrogen. Therefore 1 and 3 only are correct, option F.

1. **Find moles of $\displaystyle \ce{BaSO4}$ formed:** $\displaystyle M_r(\ce{BaSO4}) = 137.0 + 32.0 + 4(16.0) = 233.0\text{ g mol}^{-1}$ $\displaystyle n(\ce{BaSO4}) = \dfrac{0.466\text{ g}}{233.0\text{ g mol}^{-1}} = 0.00200\text{ mol}$ Each mole of $\displaystyle \ce{BaSO4}$ corresponds to $\displaystyle 1\text{ mol}$ of $\displaystyle \ce{H2SO4}$ , so in a $\displaystyle 50.0\text{ cm}^3$ sample: $\displaystyle n(\ce{H2SO4}) = 0.00200\text{ mol}$ 2. **Find moles of $\displaystyle \ce{H+}$ from $\displaystyle \ce{H2SO4}$ :**
Because sulfuric acid is diprotic: $\displaystyle n(\ce{H+})_{\text{from }\ce{H2SO4}} = 2 \times 0.00200\text{ mol} = 0.00400\text{ mol}$ 3. **Find total moles of $\displaystyle \ce{H+}$ neutralized by $\displaystyle \ce{NaOH}$ :** $\displaystyle n(\ce{OH-}) = 0.0350\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.00700\text{ mol}$ At complete neutralization, $\displaystyle n(\ce{H+})_{\text{total}} = n(\ce{OH-}) = 0.00700\text{ mol}$ .

4. **Find moles and concentration of $\displaystyle \ce{HCl}$ :** $\displaystyle n(\ce{H+})_{\text{from }\ce{HCl}} = n(\ce{H+})_{\text{total}} - n(\ce{H+})_{\text{from }\ce{H2SO4}} = 0.00700 - 0.00400 = 0.00300\text{ mol}$ Since $\displaystyle \ce{HCl}$ is monoprotic, $\displaystyle n(\ce{HCl}) = 0.00300\text{ mol}$ in $\displaystyle 50.0\text{ cm}^3$ ( $\displaystyle 0.0500\text{ dm}^3$ ): $\displaystyle [\ce{HCl}] = \dfrac{0.00300\text{ mol}}{0.0500\text{ dm}^3} = 0.060\text{ mol dm}^{-3}$

The general equation for the reaction of a Group 1 carbonate ( $\displaystyle \ce{M2CO3}$ ) with hydrochloric acid is:The stoichiometry shows that $\displaystyle 1$ mol of metal carbonate produces $\displaystyle 1$ mol of carbon dioxide gas.

Since the amount (in moles) of the carbonate reactant is identical ( $\displaystyle 0.050$ mol) in all three experiments and the acid is in excess, the amount of $\displaystyle \ce{CO2}$ produced is identical ( $\displaystyle 0.050$ mol) in each case.

At constant temperature and pressure, the volume of a gas is directly proportional to its amount in moles (Avogadro's Law). Therefore, the volumes must be equal: $\displaystyle V_1 = V_2 = V_3$ .

The molar mass of the cation or the reactivity of the metal does not affect the molar stoichiometry of the reaction.

Since the ions are isoelectronic, they all have the same number of electrons, $\displaystyle e$ . We can determine the number of protons (atomic number) for each element by considering the charge:

* $\displaystyle \mathrm{X}^{2-}$ has gained 2 electrons: Atomic number $\displaystyle = e - 2$ .

* $\displaystyle \mathrm{Y}^{-}$ has gained 1 electron: Atomic number $\displaystyle = e - 1$ .

* $\displaystyle \mathrm{Z}^{2+}$ has lost 2 electrons: Atomic number $\displaystyle = e + 2$ .

The order of atomic numbers is $\displaystyle \mathrm{X} < \mathrm{Y} < \mathrm{Z}$ .

Since X and Y form anions to reach the stable electron configuration, they belong to Groups 16 and 17 of a period (say, period $\displaystyle n$ ). Z forms a cation to reach the same configuration, meaning it must be in Group 2 of the next period ( $\displaystyle n+1$ ).

Therefore, element Z is in the period immediately following elements X and Y.

In aqueous electrolysis, there is a competition at the cathode between the reduction of the metal cation ( $\displaystyle M^{n+} + ne^- \rightarrow M$ ) and the reduction of water ( $\displaystyle 2\ce{H2O} + 2e^- \rightarrow \ce{H2} + 2\ce{OH-}$ ).
The reaction that occurs is the one with the more positive standard electrode potential ( $\displaystyle E^{\circ}$ ). In terms of the reactivity series:
* Metals more reactive than hydrogen (e.g., Na, Ca, Mg, Al) have very negative reduction potentials. Water is reduced in preference to these ions, producing hydrogen gas. * Metals less reactive than hydrogen (e.g., Cu, Ag) have positive reduction potentials. These ions are reduced in preference to water, depositing the metal.
Analyzing the list:
1. $\displaystyle \ce{AgNO3}$ : Ag is less reactive than H. Silver deposits. 2. $\displaystyle \ce{Ca(NO3)2}$ : Ca is more reactive than H. Hydrogen evolves. 3. $\displaystyle \ce{CuSO4}$ : Cu is less reactive than H. Copper deposits. 4. $\displaystyle \ce{NaCl}$ : Na is more reactive than H. Hydrogen evolves.
Thus, the metal is deposited in experiments 1 and 3 only.

First, determine the polarity of each molecule:

- $\displaystyle \text{BF}_3$ is trigonal planar and symmetrical, so it is non-polar (only London dispersion forces).
- $\displaystyle \text{BCl}_3$ is trigonal planar and symmetrical, so it is non-polar (only London dispersion forces).
- $\displaystyle \text{CH}_3\text{Cl}$ is tetrahedral (distorted) and asymmetrical, so it is polar (dipole-dipole forces + London dispersion forces).

Next, compare the strength of the intermolecular forces:

1. ** $\displaystyle \text{BF}_3$ vs $\displaystyle \text{BCl}_3$ **: Both are non-polar, but $\displaystyle \text{BCl}_3$ has significantly more electrons ( $\displaystyle 56$ vs $\displaystyle 32$ ) and a larger surface area than $\displaystyle \text{BF}_3$ . Therefore, the London dispersion forces in $\displaystyle \text{BCl}_3$ are much stronger. $\displaystyle \text{BF}_3$ has the lowest boiling point ( $\displaystyle -100^\circ\text{C}$ ).

2. ** $\displaystyle \text{CH}_3\text{Cl}$ vs $\displaystyle \text{BCl}_3$ **: Although $\displaystyle \text{CH}_3\text{Cl}$ is polar, $\displaystyle \text{BCl}_3$ is a much larger molecule with three chlorine atoms compared to one. The substantial increase in London dispersion forces in $\displaystyle \text{BCl}_3$ outweighs the dipole-dipole interactions in the smaller $\displaystyle \text{CH}_3\text{Cl}$ molecule.

Thus, $\displaystyle \text{CH}_3\text{Cl}$ ( $\displaystyle -24^\circ\text{C}$ ) lies between the two boron halides.

The correct order is $\displaystyle \text{BF}_3 < \text{CH}_3\text{Cl} < \text{BCl}_3$ .

Both have molecular formula $\displaystyle \ce{C4H10}$ but different atom connectivity, so they are structural isomers.

1. Calculate the heat energy absorbed by the water: $\displaystyle q_{\text{water}} = mc\Delta T = 600\text{ g} \times 4.2\text{ J g}^{-1\circ}\text{C}^{-1} \times 25^{\circ}\text{C} = 63\,000\text{ J} = 63.0\text{ kJ}$ 2. Calculate the total energy released by the reaction accounting for the $\displaystyle 75\%$ efficiency: $\displaystyle q_{\text{released}} = \dfrac{q_{\text{water}}}{0.75} = \dfrac{63.0\text{ kJ}}{0.75} = 84.0\text{ kJ}$ 3. Determine the enthalpy change of reaction from the energy profile: $\displaystyle \Delta H = E_{\text{products}} - E_{\text{reactants}} = -200\text{ kJ mol}^{-1} - (+80\text{ kJ mol}^{-1}) = -280\text{ kJ mol}^{-1}$ According to the stoichiometric equation: $\displaystyle \ce{2SO2(g) + O2(g) -> 2SO3(g)} \quad \Delta H = -280\text{ kJ mol}^{-1}$ This means $\displaystyle 280\text{ kJ}$ is released per $\displaystyle 2\text{ moles}$ of $\displaystyle \ce{SO2}$ reacted, or $\displaystyle 140\text{ kJ}$ per mole of $\displaystyle \ce{SO2}$ .

4. Calculate the moles of $\displaystyle \ce{SO2}$ reacted: $\displaystyle n(\ce{SO2}) = \dfrac{84.0\text{ kJ}}{140\text{ kJ mol}^{-1}} = 0.60\text{ mol}$ 5. Calculate the mass of $\displaystyle \ce{SO2}$ : $\displaystyle M_{\text{r}}(\ce{SO2}) = 32.0 + 2 \times 16.0 = 64.0\text{ g mol}^{-1}$ $\displaystyle \text{mass} = 0.60\text{ mol} \times 64.0\text{ g mol}^{-1} = 38.4\text{ g}$

Copper(II) sulfate dissolves in water but sand does not. Filtration separates sand; crystallisation recovers the dissolved salt.

To determine the mean bond enthalpy of the $\displaystyle \ce{C=O}$ bond, consider the bonds broken and bonds formed during the reaction:

Bonds broken:
- In propanone: $\displaystyle 6 \times \ce{C-H}$ , $\displaystyle 2 \times \ce{C-C}$ , $\displaystyle 1 \times \ce{C=O}$ - In $\displaystyle \ce{H2}$ : $\displaystyle 1 \times \ce{H-H}$ Bonds formed:
- In propan-2-ol: $\displaystyle 7 \times \ce{C-H}$ ( $\displaystyle 6$ on methyl groups and $\displaystyle 1$ on the central $\displaystyle \ce{CH}$ ), $\displaystyle 2 \times \ce{C-C}$ , $\displaystyle 1 \times \ce{C-O}$ , $\displaystyle 1 \times \ce{O-H}$ Cancelling bonds that remain unchanged on both sides leaves the net changes:
- Net bonds broken: $\displaystyle 1 \times \ce{C=O} + 1 \times \ce{H-H} = E(\ce{C=O}) + 436\text{ kJ mol}^{-1}$ - Net bonds formed: $\displaystyle 1 \times \ce{C-H} + 1 \times \ce{C-O} + 1 \times \ce{O-H} = 413 + 358 + 464 = 1235\text{ kJ mol}^{-1}$ Using $\displaystyle \Delta H = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$ : $\displaystyle -54 = [E(\ce{C=O}) + 436] - 1235$ $\displaystyle -54 = E(\ce{C=O}) - 799$ $\displaystyle E(\ce{C=O}) = 799 - 54 = 745\text{ kJ mol}^{-1}$ Therefore, the correct option is C.

To balance the equation, we equate the number of atoms of each element on both sides of the reaction:

1. **Copper ( $\displaystyle \ce{Cu}$ ):** $\displaystyle \text{LHS} = 3 \times 3 = 9$ $\displaystyle \text{RHS} = x \implies x = 9$ 2. **Phosphorus ( $\displaystyle \ce{P}$ ):** $\displaystyle \text{LHS} = 3 \times 1 = 3$ $\displaystyle \text{RHS} = 3 \times 1 = 3 \quad \text{(already balanced)}$ 3. **Nitrogen ( $\displaystyle \ce{N}$ ):** $\displaystyle \text{LHS} = w$ $\displaystyle \text{RHS} = 2x + y = 2(9) + y = 18 + y$ $\displaystyle y = w - 18$ 4. **Hydrogen ( $\displaystyle \ce{H}$ ):** $\displaystyle \text{LHS} = w$ $\displaystyle \text{RHS} = 3(3) + 2z = 9 + 2z$ $\displaystyle z = \dfrac{w - 9}{2}$ 5. **Oxygen ( $\displaystyle \ce{O}$ ):** $\displaystyle \text{LHS} = 3w$ $\displaystyle \text{RHS} = 6x + 3(4) + y + z = 6(9) + 12 + y + z = 66 + y + z$ Substitute the expressions for $\displaystyle y$ and $\displaystyle z$ into the oxygen balance equation: $\displaystyle 3w = 66 + (w - 18) + \dfrac{w - 9}{2}$ $\displaystyle 3w = 48 + w + \dfrac{w - 9}{2}$ $\displaystyle 2w - 48 = \dfrac{w - 9}{2}$ $\displaystyle 4w - 96 = w - 9$ $\displaystyle 3w = 87 \implies w = 29$ Now calculate $\displaystyle y$ and $\displaystyle z$ : $\displaystyle y = 29 - 18 = 11$ $\displaystyle z = \dfrac{29 - 9}{2} = 10$ Thus, the coefficients are: $\displaystyle w = 29, \quad x = 9, \quad y = 11, \quad z = 10$ The sum is: $\displaystyle w + x + y + z = 29 + 9 + 11 + 10 = 59$

1. Determine the oxidation state of phosphorus in each species:
- In $\displaystyle \ce{P4}$ , the oxidation state of $\displaystyle \text{P}$ is $\displaystyle 0$ .
- In $\displaystyle \ce{PH3}$ , hydrogen has an oxidation state of $\displaystyle +1$ , so $\displaystyle \text{P}$ has an oxidation state of $\displaystyle -3$ (reduction: gain of $\displaystyle 3\text{ e}^-$ per $\displaystyle \text{P}$ atom).
- In $\displaystyle \ce{H2PO2-}$ , hydrogen is $\displaystyle +1$ and oxygen is $\displaystyle -2$ . Let $\displaystyle x$ be the oxidation state of $\displaystyle \text{P}$ : $\displaystyle x + 2(+1) + 2(-2) = -1 \implies x - 2 = -1 \implies x = +1$ (oxidation: loss of $\displaystyle 1\text{ e}^-$ per $\displaystyle \text{P}$ atom).

2. Balance electron transfer:
- Reduction to $\displaystyle \ce{PH3}$ requires $\displaystyle 3\text{ e}^-$ .
- Oxidation to $\displaystyle \ce{H2PO2-}$ releases $\displaystyle 1\text{ e}^-$ .
- To balance electrons, $\displaystyle 3$ phosphorus atoms must be oxidized to $\displaystyle \ce{H2PO2-}$ for every $\displaystyle 1$ phosphorus atom reduced to $\displaystyle \ce{PH3}$ .

Therefore, the stoichiometric ratio of hypophosphite to phosphine produced is $\displaystyle e : d = 3 : 1$ .

(The fully balanced equation is $\displaystyle \ce{P4 + 3OH- + 3H2O -> PH3 + 3H2PO2-}$ , giving $\displaystyle a=1, b=3, c=3, d=1, e=3$ .)