Stoichiometry, Balancing Equations and the Limiting Reagent

Stoichiometry — Reading a Balanced Equation

Stoichiometry lets you predict how much product a reaction gives, or how much reactant it needs, from the balanced equation alone. The word comes from the Greek stoicheion (element) and metron (measure).

Key Point (Definition): Stoichiometry deals with the calculation of masses (and sometimes volumes) of the reactants and products in a chemical reaction.

Take the combustion of methane:

$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}$$

The parts of the equation

• Reactants (left): methane and dioxygen. Products (right): carbon dioxide and water.
• State symbols: (g) gas, (l) liquid, (s) solid, (aq) dissolved in water.
• Stoichiometric coefficients are the numbers before the formulae: 2 for $\mathrm{O_2}$ and $\mathrm{H_2O}$, 1 (unwritten) for $\mathrm{CH_4}$ and $\mathrm{CO_2}$.

Key Point: Coefficients give the number of molecules, and equally the number of moles, reacting or formed.

The same equation, read four ways

Coefficients are a ratio, so the equation can be read at any scale:

Two checks:

1. Mass is conserved: 16 + 64 = 80 g in, 44 + 36 = 80 g out.
2. Volume need not be. Here 68.1 L gives 68.1 L by coincidence; in $\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)}$, 3 volumes become 2. Volume ratios follow moles (Avogadro's law), not mass.

The volume row works only for gases at the same temperature and pressure. Molar volume is 22.7 L at STP (1 bar, 273.15 K); many JEE/NEET questions still use 22.4 L (1 atm). Check which the question expects.

The coefficients carry the ratio

The mass row is always coefficient $\times$ molar mass: 16 g of methane is one molar mass, 64 g of oxygen is two. The equation says moles react in the ratio 1 : 2 : 1 : 2, never that equal masses react.

[JEE Main] "16 g of methane needs how much oxygen?" 64 g, not 32 g, because the coefficient of $\mathrm{O_2}$ is 2.

Balancing Chemical Equations

An equation is balanced when each element has the same number of atoms on both sides: conservation of mass in shorthand.

The one rule never to break

Key Point: Change only the coefficients in front of formulae, never the subscripts. Writing $\mathrm{H_2O_2}$ to balance oxygen in the formation of water turns water into hydrogen peroxide, a different compound.

Balancing by inspection

Iron and oxygen: $\mathrm{Fe + O_2 \rightarrow Fe_2O_3}$. O is 2 per $\mathrm{O_2}$ and 3 per $\mathrm{Fe_2O_3}$; the LCM is 6, so $\mathrm{3O_2}$ and $\mathrm{2Fe_2O_3}$. That gives 4 Fe on the right, so 4Fe:

$$\mathrm{4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)}$$

Check: Fe 4 = 4, O 6 = 6.

Magnesium and oxygen: $\mathrm{Mg + O_2 \rightarrow MgO}$. Two O need 2 MgO, which needs 2 Mg:

$$\mathrm{2Mg(s) + O_2(g) \rightarrow 2MgO(s)}$$

Phosphorus and oxygen: $\mathrm{P_4 + O_2 \rightarrow P_4O_{10}}$. P is already 4 = 4; ten O need five $\mathrm{O_2}$:

$$\mathrm{P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s)}$$

The step method: combustion of propane

Step 1. Skeleton equation: $$\mathrm{C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)}$$

Step 2. Carbon: 3 C gives 3 $\mathrm{CO_2}$: $$\mathrm{C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O}$$

Step 3. Hydrogen: 8 H, 2 per $\mathrm{H_2O}$, so 4 $\mathrm{H_2O}$: $$\mathrm{C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O}$$

Step 4. Oxygen last: $3 \times 2 + 4 \times 1 = 10$ O on the right, so 5 $\mathrm{O_2}$: $$\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)}$$

Step 5. Verify. C 3 = 3, H 8 = 8, O 10 = 10.

Why oxygen goes last

A free element ($\mathrm{O_2}$, $\mathrm{H_2}$, $\mathrm{N_2}$, Fe, Mg) can be set without disturbing anything else, so balance it last; balance elements that appear in only one compound per side first.

If the last element needs a fraction (say $\frac{13}{2}\mathrm{O_2}$ for butane), finish, then double the whole equation. Fractions are fine in thermochemistry; stoichiometry uses smallest whole numbers.

A balancing checklist

[NEET] Sum of coefficients for $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$: $1 + 5 + 3 + 4 = 13$.

Stoichiometric Calculations — The Conversion Chain

Every stoichiometry problem runs on the same three-link chain.

The chain

$$\text{mass of A} \xrightarrow{\ \div M_A\ } \text{moles of A} \xrightarrow{\ \text{coefficient ratio}\ } \text{moles of B} \xrightarrow{\ \times M_B\ } \text{mass of B}$$

1. Convert the given quantity to moles: mass $\div$ molar mass, STP gas volume $\div$ 22.7 L $\mathrm{mol^{-1}}$, or (solutions, Section 9) molarity $\times$ litres.
2. Use the coefficient ratio to get moles of the wanted substance. This is the only step that uses the equation.
3. Convert to what is asked: $\times$ molar mass for grams, $\times$ 22.7 L for STP volume, $\times\ 6.022 \times 10^{23}$ for molecules.

Key Point: The coefficient ratio applies to moles only, never to grams. 1 mol $\mathrm{CH_4}$ : 2 mol $\mathrm{H_2O}$ is correct; "16 g $\mathrm{CH_4}$ : 2 g $\mathrm{H_2O}$" is meaningless.

The mole ratio as a unit factor

From $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$:

$$\frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ CH_4}} = 1 \qquad \frac{1\ \mathrm{mol\ CO_2}}{1\ \mathrm{mol\ CH_4}} = 1 \qquad \frac{2\ \mathrm{mol\ O_2}}{1\ \mathrm{mol\ CH_4}} = 1$$

Each factor equals 1, so it changes nothing physically but swaps the unit. Water from 16 g of methane:

$$16\ \mathrm{g\ CH_4} \times \frac{1\ \mathrm{mol\ CH_4}}{16\ \mathrm{g\ CH_4}} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ CH_4}} \times \frac{18\ \mathrm{g\ H_2O}}{1\ \mathrm{mol\ H_2O}} = 36\ \mathrm{g\ H_2O}$$

If the units do not cancel in pairs, a factor is upside down.

The three common problem types

Gases at the same T and P have volumes in the mole ratio: from $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$, 10 L of propane needs 50 L of oxygen and gives 30 L of $\mathrm{CO_2}$.

A problem through the chain

How many moles of methane are required to produce 22 g of $\mathrm{CO_2}$ after combustion?

• Link 1: 22 g $\mathrm{CO_2} \div 44\ \mathrm{g\ mol^{-1}} = 0.5$ mol.
• Link 2: $\mathrm{CH_4 : CO_2} = 1 : 1$, so 0.5 mol $\mathrm{CH_4}$.
• Link 3: moles are asked, so stop.

Answer: 0.5 mol of methane ($0.5 \times 16 = 8$ g).

For a sequence A to B to C, chain the ratios: moles of C = moles of A $\times$ (ratio 1) $\times$ (ratio 2); intermediate masses are not needed.

[Board] Show the balanced equation, molar masses and mole ratio; a bare "36 g" earns little.

The Limiting Reagent

Reactants are rarely supplied in exactly the proportion the equation demands, and that changes which number controls the answer.

The sandwich picture

A sandwich needs 2 slices of bread and 1 of cheese. With 10 bread and 3 cheese you make 3 sandwiches, not 5: the cheese runs out and extra bread does not help. Cheese is the limiting ingredient, bread is in excess (4 slices left). In a reaction, the reactant that runs out first stops it; the rest stays unreacted.

Key Point (Definition): When reactant amounts differ from those required by the balanced equation, the reactant present in the least amount (relative to its coefficient) is consumed first, after which no further reaction occurs whatever the amount of the other reactant. It limits the amount of product and is called the limiting reagent; the other reactant(s) are in excess.

Two cautions:

1. "Least amount" is relative to what the equation needs, not the smallest mass or mole count. 3 mol $\mathrm{H_2}$ exceeds 1 mol $\mathrm{N_2}$, yet in $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$ neither is limiting.
2. Product is always calculated from the limiting reagent; using the excess reagent gives too large an answer.

Spotting the limiting reagent

Method A (required versus available). From one reactant, use the coefficient ratio to find how much of the other it requires; if that exceeds what is available, the other reactant is limiting.

Method B (divide by the coefficient). For each reactant compute

$$\frac{\text{moles available}}{\text{stoichiometric coefficient}}$$

The smallest quotient marks the limiting reagent and is the number of times the equation runs, so moles of any product = (smallest quotient) $\times$ (product's coefficient).

In row 3, hydrogen has the larger mole count and is still limiting, because each run uses three of them.

Excess reagent left over

Leftover = available minus consumed, where consumed follows the coefficient ratio. In row 4, 1 mol $\mathrm{N_2}$ consumes 3 mol $\mathrm{H_2}$, leaving $6 - 3 = 3$ mol.

[JEE Main] When reactants are given in grams, convert both to moles before dividing by coefficients. 10 kg of hydrogen sounds like less than 50 kg of nitrogen, but the coefficient 3 makes hydrogen the limiting reagent, as the next block shows.

Ammonia Synthesis — A Limiting-Reagent Problem in Full

The Haber process, $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$, is the standard limiting-reagent problem.

Problem 1.5. 50.0 kg of $\mathrm{N_2(g)}$ and 10.0 kg of $\mathrm{H_2(g)}$ are mixed to produce $\mathrm{NH_3(g)}$. Calculate the amount of $\mathrm{NH_3(g)}$ formed. Identify the limiting reagent.

Step 1: Equation and molar masses

$$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$$

$M(\mathrm{N_2}) = 28.0\ \mathrm{g\ mol^{-1}}$, $M(\mathrm{H_2}) = 2.016\ \mathrm{g\ mol^{-1}}$, $M(\mathrm{NH_3}) = 17.0\ \mathrm{g\ mol^{-1}}$.

Step 2: Reactants to moles

$$n_{\mathrm{N_2}} = 50.0\ \mathrm{kg} \times \frac{1000\ \mathrm{g}}{1\ \mathrm{kg}} \times \frac{1\ \mathrm{mol}}{28.0\ \mathrm{g}} = 1786\ \mathrm{mol} = 17.86 \times 10^{2}\ \mathrm{mol}$$

$$n_{\mathrm{H_2}} = 10.0\ \mathrm{kg} \times \frac{1000\ \mathrm{g}}{1\ \mathrm{kg}} \times \frac{1\ \mathrm{mol}}{2.016\ \mathrm{g}} = 4960\ \mathrm{mol} = 4.96 \times 10^{3}\ \mathrm{mol}$$

Step 3: Limiting reagent

Method A: 1786 mol $\mathrm{N_2}$ would need

$$17.86 \times 10^{2}\ \mathrm{mol\ N_2} \times \frac{3\ \mathrm{mol\ H_2}}{1\ \mathrm{mol\ N_2}} = 5.36 \times 10^{3}\ \mathrm{mol\ H_2}$$

Only $4.96 \times 10^{3}$ mol is available, so dihydrogen is limiting and dinitrogen is in excess.

Method B: $\mathrm{N_2}$: $1786/1 = 1786$; $\mathrm{H_2}$: $4960/3 = 1653$. Smaller quotient, so hydrogen limits.

Step 4: Product from the limiting reagent

$$4.96 \times 10^{3}\ \mathrm{mol\ H_2} \times \frac{2\ \mathrm{mol\ NH_3}}{3\ \mathrm{mol\ H_2}} = 3.30 \times 10^{3}\ \mathrm{mol\ NH_3}$$

Step 5: Mass

$$3.30 \times 10^{3}\ \mathrm{mol} \times 17.0\ \mathrm{g\ mol^{-1}} = 56.1 \times 10^{3}\ \mathrm{g} = 56.1\ \mathrm{kg\ NH_3}$$

Answer: $3.30 \times 10^{3}$ mol, i.e. 56.1 kg of ammonia; $\mathrm{H_2}$ is the limiting reagent.

The follow-up examiners ask

Nitrogen consumed $= 4960\ \mathrm{mol\ H_2} \times \frac{1}{3} = 1653$ mol. Left $= 1786 - 1653 = 133$ mol $= 133 \times 28.0 = 3.72 \times 10^{3}$ g, about 3.7 kg.

Mass check: 50.0 + 10.0 = 60.0 kg in; 56.1 kg $\mathrm{NH_3}$ + 3.7 kg $\mathrm{N_2}$ = 59.8 kg out, equal within rounding.

Hydrogen was 10 kg against 50 kg of nitrogen and still limiting: small molar mass (2.016), large coefficient (3). Only moles divided by coefficient decides.

[JEE Main] Variants: swap the masses (50 kg $\mathrm{H_2}$ + 10 kg $\mathrm{N_2}$ makes $\mathrm{N_2}$ limiting); give moles; ask for the leftover; or ask what mass of $\mathrm{H_2}$ exactly consumes 50.0 kg of $\mathrm{N_2}$ (1786 mol $\times$ 3 $\times$ 2.016 g = 10.8 kg).

Stoichiometry in Solutions and the Exam Checklist

A reactant in solution

For a dissolved reactant, given as molarity and volume,

$$n = M \times V\ (\text{in litres})$$

Two exercises belong here because they are pure stoichiometry once the moles are known (concentration itself is Section 9).

Exercise 1.35. What mass of $\mathrm{CaCO_3}$ is required to react completely with 25 mL of 0.75 M HCl?

$$\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}$$

$n(\mathrm{HCl}) = 0.75\ \mathrm{mol\ L^{-1}} \times 0.025\ \mathrm{L} = 0.01875$ mol. Ratio 1 : 2, so $0.009375$ mol $\mathrm{CaCO_3}$; with $M = 100.1\ \mathrm{g\ mol^{-1}}$, mass $= 0.009375 \times 100.1 = 0.938$ g.

Exercise 1.36. How many grams of HCl react with 5.0 g of $\mathrm{MnO_2}$? $\mathrm{4HCl + MnO_2 \rightarrow 2H_2O + MnCl_2 + Cl_2}$. $n(\mathrm{MnO_2}) = 5.0/86.9 = 0.0575$ mol; HCl $= 4 \times 0.0575 = 0.230$ mol $= 0.230 \times 36.5 = 8.40$ g.

Only link 1 of the chain is new.

Gas-volume stoichiometry

For a gaseous product at STP, link 3 is $\times 22.7$ L $\mathrm{mol^{-1}}$: 8 g of methane (0.5 mol) gives 0.5 mol $\mathrm{CO_2} = 0.5 \times 22.7 = 11.35$ L. If all gases are at the same T and P, coefficients give the volume ratio directly (Gay Lussac, Section 4).

The five errors that cost the most marks

A 30-second recipe

1. Balance the equation.
2. Moles of everything given (mass $\div M$; volume $\div 22.7$; $M \times V$).
3. Divide each reactant's moles by its coefficient; smallest is limiting (skip if only one reactant is quantified).
4. Smallest quotient $\times$ product's coefficient = moles of product.
5. Convert to grams, litres, molecules, or leftover excess as asked.

Key Point: Balanced equation, then moles, then ratio, then answer.

[NEET] NEET stoichiometry is usually one mole ratio with clean numbers (4.4 g $\mathrm{CO_2}$, 5.6 L at STP, 3.4 g $\mathrm{NH_3}$); if you are dividing by 2.016, check whether $M(\mathrm{H_2}) = 2$ is intended. JEE Main adds per cent yield and purity on top of the theoretical yield.

Solved Examples

Question 1: Reading a balanced equation four ways

For $\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)}$, state what the equation tells you in terms of (i) molecules, (ii) moles, (iii) volumes at STP, and (iv) masses. Then check mass conservation.

Answer: Molecules: 2 $\mathrm{H_2}$ + 1 $\mathrm{O_2}$ give 2 $\mathrm{H_2O}$. Moles: 2 mol + 1 mol give 2 mol.

Volumes at STP: $2 \times 22.7 = 45.4$ L $\mathrm{H_2}$ + 22.7 L $\mathrm{O_2}$ give 45.4 L of water vapour. 68.1 L becomes 45.4 L; volume is not conserved.

Masses: $2 \times 2.016 = 4.03$ g $\mathrm{H_2}$ + 32.0 g $\mathrm{O_2}$ give $2 \times 18.0 = 36.0$ g $\mathrm{H_2O}$. Check: 4.03 + 32.0 = 36.03 g in, 36.0 g out, equal within rounding.

Ans: 2 : 1 : 2 in molecules and moles; 45.4 L : 22.7 L : 45.4 L at STP; 4.03 g : 32.0 g : 36.0 g.

Question 2: Balancing by inspection

Balance: (i) $\mathrm{Fe + O_2 \rightarrow Fe_2O_3}$ (ii) $\mathrm{P_4 + O_2 \rightarrow P_4O_{10}}$ (iii) $\mathrm{Al + O_2 \rightarrow Al_2O_3}$.

Answer: (i) O is 2 on the left, 3 on the right; LCM 6 gives $\mathrm{3O_2}$ and $\mathrm{2Fe_2O_3}$, then 4 Fe on the right needs $\mathrm{4Fe}$. $\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}$. Check: Fe 4 = 4, O 6 = 6.

(ii) P is 4 = 4 already; ten O needs $\mathrm{5O_2}$. $\mathrm{P_4 + 5O_2 \rightarrow P_4O_{10}}$. Check: O 10 = 10.

(iii) Same as iron: $\mathrm{4Al + 3O_2 \rightarrow 2Al_2O_3}$. Check: Al 4 = 4, O 6 = 6.

Ans: (i) $\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}$; (ii) $\mathrm{P_4 + 5O_2 \rightarrow P_4O_{10}}$; (iii) $\mathrm{4Al + 3O_2 \rightarrow 2Al_2O_3}$.

Question 3: The step method with a fractional coefficient

Balance the combustion of butane, $\mathrm{C_4H_{10} + O_2 \rightarrow CO_2 + H_2O}$, using the C, H, O order.

Answer: 4 C gives $\mathrm{4CO_2}$; 10 H gives $\mathrm{5H_2O}$.

Oxygen last: $4 \times 2 + 5 \times 1 = 13$ O on the right needs $\frac{13}{2}\mathrm{O_2}$: $\mathrm{C_4H_{10} + \tfrac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O}$.

Doubling every coefficient clears the fraction: $\mathrm{2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O}$. Verify: C 8 = 8, H 20 = 20, O 26 = 16 + 10 = 26.

Ans: $\mathrm{2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l)}$.

Watch out: Never fix the fraction by changing $\mathrm{O_2}$ to O; double the whole equation.

Question 4: Water from 16 g of methane

Calculate the amount of water (in g) produced by the combustion of 16 g of methane.

Answer: $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}$.

$M(\mathrm{CH_4}) = 12 + 4 \times 1 = 16\ \mathrm{g\ mol^{-1}}$, so 16 g is 1 mol, which gives 2 mol $\mathrm{H_2O}$ by the 1 : 2 ratio.

$M(\mathrm{H_2O}) = 2 + 16 = 18\ \mathrm{g\ mol^{-1}}$; mass $= 2 \times 18 = 36$ g.

In one line: $16\ \mathrm{g\ CH_4} \times \frac{1\ \mathrm{mol\ CH_4}}{16\ \mathrm{g}} \times \frac{2\ \mathrm{mol\ H_2O}}{1\ \mathrm{mol\ CH_4}} \times \frac{18\ \mathrm{g}}{1\ \mathrm{mol\ H_2O}} = 36\ \mathrm{g}$. Oxygen is in excess since no amount is given.

Ans: 36 g of water.

Question 5: Methane needed for 22 g of carbon dioxide

How many moles of methane are required to produce 22 g of $\mathrm{CO_2(g)}$ after combustion?

Answer: From $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}$, $\mathrm{CH_4 : CO_2} = 1 : 1$.

$n(\mathrm{CO_2}) = 22\ \mathrm{g} \times \frac{1\ \mathrm{mol}}{44\ \mathrm{g}} = 0.5$ mol, so 0.5 mol $\mathrm{CH_4}$.

Check by mass: 44 g $\mathrm{CO_2}$ needs 16 g $\mathrm{CH_4}$, so 22 g needs 8 g $= 8/16 = 0.5$ mol.

Ans: 0.5 mol of methane (8 g).

Question 6: Mole-to-volume — gas volumes at STP

8.0 g of methane is burnt completely in oxygen. Calculate (i) the volume of $\mathrm{O_2}$ at STP required, (ii) the volume of $\mathrm{CO_2}$ at STP produced, and (iii) the volume of $\mathrm{O_2}$ needed if the gases are measured at the same room temperature and pressure and the methane occupies 2.0 L.

Answer: $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$; $n(\mathrm{CH_4}) = 8.0/16 = 0.50$ mol.

(i) $0.50 \times 2 = 1.0$ mol $\mathrm{O_2}$; at STP (1 bar) $1.0 \times 22.7 = 22.7$ L (22.4 L at 1 atm).

(ii) 0.50 mol $\mathrm{CO_2} = 0.50 \times 22.7 = 11.35$ L, about 11.4 L.

(iii) Same T and P, so volumes follow coefficients: 2.0 L $\mathrm{CH_4}$ needs $2 \times 2.0 = 4.0$ L $\mathrm{O_2}$.

Ans: (i) 22.7 L; (ii) 11.4 L; (iii) 4.0 L.

Watch out: Molar volume is needed only to turn mass or moles into litres; gas to gas at the same conditions is just the coefficient ratio.

Question 7: Carbon burnt in limited oxygen

Calculate the amount of carbon dioxide produced when (i) 1 mole of carbon is burnt in air, (ii) 1 mole of carbon is burnt in 16 g of dioxygen, (iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer: $\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$, ratio 1 : 1 : 1. $M(\mathrm{O_2}) = 32$, $M(\mathrm{CO_2}) = 44\ \mathrm{g\ mol^{-1}}$.

(i) Air means unlimited oxygen: 1 mol C gives 1 mol $\mathrm{CO_2} = 44$ g.

(ii) Oxygen $= 16/32 = 0.5$ mol. C: $1/1 = 1$; $\mathrm{O_2}$: $0.5/1 = 0.5$, so oxygen limits. $\mathrm{CO_2} = 0.5$ mol $= 22$ g; 0.5 mol C (6 g) left.

(iii) Carbon 2 mol, oxygen still 0.5 mol and still limiting. $\mathrm{CO_2} = 0.5$ mol $= 22$ g; 1.5 mol C (18 g) left.

Ans: (i) 44 g; (ii) 22 g; (iii) 22 g.

Watch out: Doubling the carbon changed nothing; extra excess reagent never raises the yield.

Question 8: Limiting reagent in A + B2 → AB2

In the reaction $\mathrm{A + B_2 \rightarrow AB_2}$, identify the limiting reagent, if any, in: (i) 300 atoms of A + 200 molecules of $\mathrm{B_2}$; (ii) 2 mol A + 3 mol $\mathrm{B_2}$; (iii) 100 atoms of A + 100 molecules of $\mathrm{B_2}$; (iv) 5 mol A + 2.5 mol $\mathrm{B_2}$; (v) 2.5 mol A + 5 mol $\mathrm{B_2}$.

Answer: The ratio is 1 : 1, so the smaller count is limiting.

(i) 300 A, 200 $\mathrm{B_2}$: $\mathrm{B_2}$ limits; 100 A left.

(ii) 2 mol A, 3 mol $\mathrm{B_2}$: A limits; 1 mol $\mathrm{B_2}$ left.

(iii) 100 and 100: stoichiometric, none limiting.

(iv) 5 mol A, 2.5 mol $\mathrm{B_2}$: $\mathrm{B_2}$ limits; 2.5 mol A left.

(v) 2.5 mol A, 5 mol $\mathrm{B_2}$: A limits; 2.5 mol $\mathrm{B_2}$ left.

Ans: (i) $\mathrm{B_2}$; (ii) A; (iii) none; (iv) $\mathrm{B_2}$; (v) A.

Watch out: "Smaller amount limits" holds only for 1 : 1 coefficients; otherwise divide by the coefficient first.

Question 9: 50.0 kg N2 and 10.0 kg H2

50.0 kg of $\mathrm{N_2(g)}$ and 10.0 kg of $\mathrm{H_2(g)}$ are mixed to produce $\mathrm{NH_3(g)}$. Calculate the amount of $\mathrm{NH_3(g)}$ formed and identify the limiting reagent.

Answer: $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$.

$n_{\mathrm{N_2}} = 50.0 \times 10^{3} / 28.0 = 17.86 \times 10^{2}$ mol; $n_{\mathrm{H_2}} = 10.0 \times 10^{3} / 2.016 = 4.96 \times 10^{3}$ mol.

$17.86 \times 10^{2}$ mol $\mathrm{N_2}$ needs $3 \times 17.86 \times 10^{2} = 5.36 \times 10^{3}$ mol $\mathrm{H_2}$; only $4.96 \times 10^{3}$ mol is there, so $\mathrm{H_2}$ limits. (Check: $1786/1 = 1786$ against $4960/3 = 1653$.)

$\mathrm{NH_3} = 4.96 \times 10^{3} \times \frac{2}{3} = 3.30 \times 10^{3}$ mol; mass $= 3.30 \times 10^{3} \times 17.0 = 56.1 \times 10^{3}$ g $= 56.1$ kg.

Ans: $3.30 \times 10^{3}$ mol $= 56.1$ kg of $\mathrm{NH_3}$; dihydrogen is limiting.

Watch out: The smaller mass limits here only because $4960/3 < 1786/1$; always compare moles over coefficients.

Question 10: 2.00 × 10^3 g N2 with 1.00 × 10^3 g H2

Dinitrogen and dihydrogen react to produce ammonia. (i) Calculate the mass of ammonia produced if $2.00 \times 10^{3}$ g of dinitrogen reacts with $1.00 \times 10^{3}$ g of dihydrogen. (ii) Will any reactant remain unreacted? (iii) If yes, which one and what is its mass?

Answer: The given $\mathrm{N_2 + H_2 \rightarrow 2NH_3}$ is unbalanced; I use $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$.

$n_{\mathrm{N_2}} = 2000/28.0 = 71.4$ mol; $n_{\mathrm{H_2}} = 1000/2.016 = 496$ mol. $71.4/1 = 71.4$ against $496/3 = 165$, so $\mathrm{N_2}$ limits.

(i) $71.4 \times 2 = 142.9$ mol $\mathrm{NH_3}$; mass $= 142.9 \times 17.0 = 2.43 \times 10^{3}$ g.

(ii) Yes, dihydrogen remains.

(iii) $\mathrm{H_2}$ used $= 71.4 \times 3 = 214.3$ mol $= 214.3 \times 2.016 = 432$ g; left $= 1000 - 432 = 568$ g (571 g with $M(\mathrm{H_2}) = 2.0$).

Mass check: $2430 + 568 = 2998 \approx 3000$ g.

Ans: (i) $2.43 \times 10^{3}$ g of $\mathrm{NH_3}$; (ii) yes; (iii) about 568 g of $\mathrm{H_2}$.

Watch out: The equation in the question was not balanced. Check before using it.

Question 11: CaCO3 needed for 25 mL of 0.75 M HCl

Calcium carbonate reacts with aqueous HCl according to $\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}$. What mass of $\mathrm{CaCO_3}$ is required to react completely with 25 mL of 0.75 M HCl?

Answer: $n(\mathrm{HCl}) = M \times V = 0.75 \times \frac{25}{1000} = 0.01875$ mol.

2 mol HCl per mol $\mathrm{CaCO_3}$, so $n(\mathrm{CaCO_3}) = 0.01875/2 = 0.009375$ mol.

$M(\mathrm{CaCO_3}) = 40.1 + 12.0 + 3 \times 16.0 = 100.1\ \mathrm{g\ mol^{-1}}$; mass $= 0.009375 \times 100.1 = 0.938$ g (0.9375 g with $M = 100$).

If asked: $\mathrm{CO_2} = 0.009375 \times 22.7 = 0.213$ L $= 213$ mL at STP.

Ans: About 0.94 g of $\mathrm{CaCO_3}$.

Watch out: The 2 : 1 ratio halves the HCl moles, not doubles.

Question 12: HCl consumed by 5.0 g of MnO2

Chlorine is prepared by $\mathrm{4HCl(aq) + MnO_2(s) \rightarrow 2H_2O(l) + MnCl_2(aq) + Cl_2(g)}$. How many grams of HCl react with 5.0 g of manganese dioxide? Also find the volume of $\mathrm{Cl_2}$ at STP.

Answer: $M(\mathrm{MnO_2}) = 54.9 + 2 \times 16.0 = 86.9\ \mathrm{g\ mol^{-1}}$; $n(\mathrm{MnO_2}) = 5.0/86.9 = 0.0575$ mol.

$n(\mathrm{HCl}) = 4 \times 0.0575 = 0.230$ mol; with $M(\mathrm{HCl}) = 1.0 + 35.5 = 36.5\ \mathrm{g\ mol^{-1}}$, mass $= 0.230 \times 36.5 = 8.40$ g.

$\mathrm{Cl_2}$ is 1 : 1 with $\mathrm{MnO_2}$: $0.0575 \times 22.7 = 1.31$ L at STP (1.29 L at 1 atm, 22.4 L).

Ans: 8.40 g of HCl; about 1.3 L of $\mathrm{Cl_2}$ at STP.

Watch out: Coefficient 4 on HCl multiplies the $\mathrm{MnO_2}$ moles by 4; write the ratio as a fraction so units cancel.