Worked examples — Important compounds — NaOH, NaCl, Na₂CO₃ (Solvay), NaHCO₃; CaO, CaCO₃, gypsum, plaster of Paris

Everything below uses only these atomic masses (g mol⁻¹), so keep them in front of you:

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The scenario matrix

Every question in this topic is one (or a mix) of these eight cells (C1–C8). The last column tells you which worked example covers it.

#	Case class	What makes it tricky	Covered by
C1	Mass → mass along one reaction	pick the right molar-mass ratio	Ex 1
C2	Mass → gas volume (STP)	must use 22.4 L mol⁻¹	Ex 2
C3	Limiting reagent (two reactants given)	one runs out first	Ex 3
C4	Hydrate water-of-crystallisation	count only the water lost	Ex 4
C5	Multi-step / Solvay chain yield	product of two conversions	Ex 5
C6	Degenerate / limiting-value case (over-heating, excess/deficient gas)	reaction switches product	Ex 6
C7	Equilibrium / Le Chatelier direction	no numbers, pure reasoning	Ex 7
C8	Real-world word problem (purity)	hidden impurity fraction	Ex 8

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Ex 1 — C1: Mass → mass (calcination of limestone)

Forecast: guess now — more than 50 g, exactly 50 g, or less? (CO₂ leaves, so...)

1. Find molar masses. Why this step? Grams mean nothing until we convert to moles; molar mass is the exchange rate. In the bar chart the single 50-unit INPUT block becomes a 28-unit green block plus a 22-unit red block — same total height, just repartitioned into solid CaO and escaping CO₂. $M({\text{CaCO}}_3)=40+12+3(16)=100$, $M(\text{CaO})=40+16=56$.
2. Convert 50 g of CaCO₃ to moles. Why? The balanced equation speaks in moles, not grams. $n({\text{CaCO}}_3)=\frac{50}{100}=0.5\text{mol}.$
3. Use the mole ratio. Why? The equation says $1{\text{CaCO}}_3\to 1\text{CaO}$, so moles carry across 1:1. $n(\text{CaO})=0.5\text{mol}.$
4. Back to grams. Why? The question asked for mass. $m(\text{CaO})=0.5\times 56=28\text{g}.$

Verify: Mass must be conserved. CaCO₃ in = 50 g. Out = CaO (28 g) + CO₂. $n({\text{CO}}_2)=0.5$, $M({\text{CO}}_2)=44$, so CO₂ = 22 g. $28+22=50$ ✓ (this is the green + red = blue of the bar chart). Answer less than 50 g — the escaping CO₂ is the lost weight. Forecast check passes.

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Ex 2 — C2: Mass → gas volume (baking soda in the oven)

Forecast: will it be roughly a cupful (≈0.25 L), a jug (≈1 L), or a bucket (≈10 L)?

1. Molar mass of NaHCO₃. Why? Convert grams to moles first. $M=23+1+12+3(16)=84$.
2. Moles of NaHCO₃. Why? 8.4 g is exactly a tenth of 84. $n=\frac{8.4}{84}=0.1\text{mol}.$
3. Mole ratio to CO₂. Why? The equation shows 2 NaHCO₃ give 1 CO₂ — a 2:1 ratio, so halve. $n({\text{CO}}_2)=\frac{0.1}{2}=0.05\text{mol}.$
4. Moles → litres at STP. Why this tool? Because "volume of a gas" is asked, and 1 mol gas = 22.4 L is the only bridge from count to volume. On the litre number line described above, the cup sits at 0.25 L and the bucket at 10 L; our 0.05 mol lands the green answer marker at 1.12 L, comfortably between the two — a jug. $V=0.05\times 22.4=1.12\text{L}.$

Verify: 0.05 mol is a small amount, and $0.05\times 22.4=1.12$ L — a jug's worth, between the cup (0.25 L) and bucket (10 L) markers. Sensible for a cake. ✓

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Ex 3 — C3: Limiting reagent (neutralising CO₂ with NaOH)

Forecast: guess whether NaOH or CO₂ is the limiting reactant.

1. Moles of each. Why? You cannot compare "who runs out" in grams — only in moles. $M(\text{NaOH})=23+16+1=40\Rightarrow n=\frac{8}{40}=0.2\text{mol}$. $M({\text{CO}}_2)=44\Rightarrow n=\frac{4.4}{44}=0.1\text{mol}$.
2. Compare ratio to the carbonate reaction $2\text{NaOH}:1{\text{CO}}_2$. Why start here? Because we have twice as much NaOH as CO₂ ($0.2$ vs $0.1$) — exactly the 2:1 ratio the carbonate reaction wants. In the figure each red CO₂ is bonded to two blue NaOH, and nothing is left over.
3. Check the fit. Why? $0.1$ mol CO₂ needs $0.2$ mol NaOH — and we have exactly $0.2$. Neither is in excess; they finish together for the carbonate path.
4. Conclude. Why carbonate not bicarbonate? Bicarbonate forms only when CO₂ is in excess; here CO₂ is the deficient gas (just enough for 2:1). Product = ${\text{Na}}_2{\text{CO}}_3$. $n({\text{Na}}_2{\text{CO}}_3)=0.1\text{mol}.$

Verify: Product mass: $M({\text{Na}}_2{\text{CO}}_3)=2(23)+12+3(16)=106$, so $0.1\times 106=10.6$ g. Mass in = $8+4.4=12.4$ g; out = Na₂CO₃ (10.6 g) + H₂O ($0.1\times 18=1.8$ g) $=12.4$ g ✓. Neither reactant is in excess — the amounts are in the exact stoichiometric ratio, so both are consumed completely and there is no single "leftover" limiting reagent; this is the special exact-fit case.

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Ex 4 — C4: Hydrate water-of-crystallisation (washing soda → PoP logic)

Forecast: the figure shows 4 water molecules going in and 1 staying. How many leave — 1, 2 or 3?

1. Molar mass of gypsum. Why? Convert 344 g to moles. $M({\text{CaSO}}_4)=40+32+4(16)=136$; add $2\times 18=36$ for the two waters: $M=172$.
2. Moles of gypsum. $n=\frac{344}{172}=2\text{mol}$. Why 2? Convenient — the equation is written per 2 gypsum units, so we have exactly one "reaction packet".
3. Water lost. Why 3, not 4? Look at the three red arrows in the figure: two gypsum units carry $2\times 2=4$ waters; PoP keeps 1 shared water; so $4-1=3$ waters leave per packet. $n({\text{H}}_2\text{O lost})=3\text{mol}\Rightarrow m=3\times 18=54\text{g}.$
4. PoP mass by conservation. Why subtract? Nothing else leaves, so PoP = gypsum − lost water. $m(\text{PoP})=344-54=290\text{g}.$

Verify: Directly, $M(\text{PoP})=({\text{CaSO}}_4{)}_2\cdot {\text{H}}_2\text{O}=2(136)+18=290$; we made 1 mol of it → 290 g ✓. Matches the subtraction. This "½ water per CaSO₄" is exactly why PoP later re-absorbs water to set.

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Ex 5 — C5: Multi-step Solvay chain yield

Forecast: more or fewer moles of Na₂CO₃ than NaCl? (Two NaCl feed one Na₂CO₃...)

1. Moles of NaCl. $M(\text{NaCl})=23+35.5=58.5\Rightarrow n=\frac{117}{58.5}=2\text{mol}$. Why? The chain starts from NaCl.
2. NaCl → NaHCO₃ (1:1). Why? First equation: each NaCl gives one NaHCO₃. So $n({\text{NaHCO}}_3)=2\text{mol}$.
3. NaHCO₃ → Na₂CO₃ (2:1). Why halve? Calcination needs 2 bicarbonates per carbonate. $n({\text{Na}}_2{\text{CO}}_3)=\frac{2}{2}=1\text{mol}.$
4. Mass. $M({\text{Na}}_2{\text{CO}}_3)=2(23)+12+3(16)=106\Rightarrow m=1\times 106=106\text{g}$.

Verify: Sodium-atom bookkeeping: 2 mol NaCl carry 2 mol Na atoms; 1 mol Na₂CO₃ holds $2\times 1=2$ mol Na atoms ✓ — no sodium lost. Answer 106 g.

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Ex 6 — C6: Degenerate / limiting case (over-heating PoP)

Forecast: does more water leave, and does that help or ruin the setting?

1. Identify the degenerate product. Why "degenerate"? Above ~473 K all crystal water leaves, giving anhydrous ${\text{CaSO}}_4$ (dead-burnt anhydrite) — the reaction has run past its useful endpoint. $CaS{O}_4\cdot 2H_{2}O\stackrel{500K}{\to }CaS{O}_4+2H_{2}O.$
2. Moles of gypsum. $n=\frac{172}{172}=1\text{mol}$ (from Ex 4's molar mass).
3. Water lost. Why 2 now, not 1.5? All 2 waters per formula unit go. $n({\text{H}}_2\text{O})=2\text{mol}\Rightarrow 36\text{g}$.
4. Why it won't set. Why does the ½-water matter? PoP sets because its retained ½ water provides nucleation sites so it re-crystallises into interlocking gypsum. Dead-burnt CaSO₄ has zero retained water and re-hydrates far too slowly — no interlocking mass forms.

Verify: Compare with Ex 4: controlled 393 K over 2 mol gypsum lost 54 g (3 water per 2 units = 1.5 per unit → for 1 mol that's 27 g). Over-heating 1 mol here loses 36 g — indeed more water leaves ($36>27$), confirming the "dries faster" intuition, yet ruins the product. ✓

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Ex 7 — C7: Le Chatelier reasoning (no numbers)

Forecast: which "stress" does the open kiln remove?

1. Name the stress. Why? Le Chatelier says a system shifts to undo an imposed change. The only gaseous product is CO₂.
2. Open kiln removes CO₂. Why does removing a product help? Lowering $[{\text{CO}}_2]$ makes the system replace it → equilibrium shifts forward (right), decomposing more CaCO₃.
3. Sealed container. Why does it stall? CO₂ builds up, pushing the equilibrium back (left) until forward and back balance — reaction stops early.

That fully answers the open-vs-sealed question. The next callout is a bonus extension, not needed for the asked question.

Recall Optional extension: the temperature effect (not asked, but good to know)

Why does heat also drive it? Endothermic = heat behaves like a "reactant"; adding heat shifts the equilibrium forward. So in a real kiln two stresses cooperate — high temperature and CO₂ removal — but only the CO₂-removal part answers the open-vs-sealed question above.

Verify: Consistency check — the parent note states "CO₂ escapes drives it forward — Le Chatelier in an open kiln." Our reasoning matches, and no formula contradicts it. ✓ (Nothing numeric to check.)

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Ex 8 — C8: Real-world word problem (impure limestone)

Forecast: the hidden trap — do you use 250 kg or something smaller in the stoichiometry?

1. Extract the pure CaCO₃. Why? Only the CaCO₃ reacts; sand is a spectator. This is the trap — never feed impure mass into a mole ratio. $m({\text{CaCO}}_3)=250\times 0.80=200\text{kg}.$
2. Convert to moles — mind the units. Why kmol? Our molar masses are in g mol⁻¹, and $1\text{kg}=1000\text{g}$. A neat shortcut: since $M({\text{CaCO}}_3)=100{\text{g mol}}^{-1}$ is numerically the same as $100{\text{kg kmol}}^{-1}$, dividing kilograms by the molar-mass number gives kilomoles (1 kmol = 1000 mol). So $200\text{kg}÷100=2\text{kmol}$ — this is identical to working in g and mol, just scaled by 1000 on both sides. $n({\text{CaCO}}_3)=\frac{200\text{kg}}{100{\text{kg kmol}}^{-1}}=2\text{kmol}.$
3. Apply the 1:1:1 stoichiometry $CaC{O}_3\to CaO+C{O}_2$. Why? One carbonate gives exactly one oxide and one CO₂, so the kilomoles carry straight across. $n(\text{CaO})=2\text{kmol},n({\text{CO}}_2)=2\text{kmol}.$
4. Convert kilomoles back to mass. Why? The question asked for masses of quicklime and of released CO₂. As in step 2, kmol × (molar-mass number) = kg. $m(\text{CaO})=2\times 56=112\text{kg},m({\text{CO}}_2)=2\times 44=88\text{kg}.$ So the kiln yields 112 kg of quicklime and vents 88 kg of CO₂.

Verify: Mass balance on the reacting part: 200 kg CaCO₃ → 112 kg CaO + 88 kg CO₂, and $112+88=200$ ✓. The 50 kg sand passes through unchanged, so total solid out = $112+50=162$ kg. ✓

Recall Self-test: which cell does each hit?

Ex 1 hits ::: C1 (mass→mass) Ex 3 hits ::: C3 (limiting reagent — the exact-fit special case, neither reactant in excess) Ex 6 hits ::: C6 (degenerate over-heating → anhydrite) Why must Ex 8 use 200 kg not 250 kg? ::: Only the 80 % that is CaCO₃ reacts; the sand is inert.

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Related build-up: Electrolysis and Electrode Potentials for the chlor-alkali cell, Salt Hydrolysis and pH for why Na₂CO₃ is alkaline, and Hardness of Water for where CaCO₃ and Ca(OH)₂ reappear.