Limiting reagent problems: Exercises — Chemistry deep dive

Before we start, one picture to keep in your head the whole time: the funding bars. Each reactant "funds" a certain number of complete reaction runs; the shortest bar caps everyone.

Level 1 — Recognition

Goal: read off the limiting reagent when the numbers are already moles.

Exercise 1.1 (L1)

For the reaction $2 H_{2} + O_{2} \to 2 H_{2} O,$ you have $6$ mol $H_{2}$ and $2$ mol $O_{2}$ . Which reactant is limiting?

Recall Solution 1.1

Step — compute the ratio $R_{i} = n_{i} / ν_{i}$ (how many full reaction sets each can fund). $R_{H_{2}} = \frac{6}{2} = 3, R_{O_{2}} = \frac{2}{1} = 2.$ What we did: divided moles-you-have by the coefficient the recipe demands. Why: $R$ tells you how many times the reaction can run on that reactant alone. $2 < 3$ , so == $O_{2}$ is limiting==. $H_{2}$ is in excess.

Exercise 1.2 (L1)

For $N_{2} + 3 H_{2} \to 2 N H_{3},$ you have $4$ mol $N_{2}$ and $6$ mol $H_{2}$ . Which is limiting, and how many mol of the excess reactant remain?

Recall Solution 1.2

$R_{N_{2}} = \frac{4}{1} = 4, R_{H_{2}} = \frac{6}{3} = 2.$ $2 < 4 \Rightarrow$ == $H_{2}$ is limiting==. Leftover $N_{2}$ : the reaction runs $R_{H_{2}} = 2$ full sets. Each set eats $1$ mol $N_{2}$ , so $N_{2}$ used $= 2 \times 1 = 2$ mol. Leftover $= 4 - 2 = 2 mol N_{2}$ . Why subtract: the excess reactant doesn't vanish — only what pairs with the limiting reagent reacts.

Level 2 — Application

Goal: convert grams to moles yourself, then find the limiting reagent and product mass.

Exercise 2.1 (L2)

$C + O_{2} \to C O_{2}$ Burn $24 g$ of carbon in $64 g$ of $O_{2}$ . Limiting reagent and mass of $C O_{2}$ ? ( $M_{C} = 12$ , $M_{O_{2}} = 32$ , $M_{C O_{2}} = 44$ )

Recall Solution 2.1

Step 1 — moles (recipe counts particles, so leave grams): $n = m / M$ . $n_{C} = \frac{24}{12} = 2 mol, n_{O_{2}} = \frac{64}{32} = 2 mol .$ Step 2 — ratios (both coefficients are $1$ ): $R_{C} = \frac{2}{1} = 2, R_{O_{2}} = \frac{2}{1} = 2.$ They are equal — a perfect stoichiometric mix, no excess. Either can be called limiting. Step 3 — product: $n_{C O_{2}} = 2 \times \frac{1}{1} = 2$ mol $\Rightarrow m = 2 \times 44 = 88 g$ .

Exercise 2.2 (L2)

$Z n + 2 H C l \to Z n C l_{2} + H_{2}$ $13 g$ zinc reacts with $14.6 g$ HCl. Find the limiting reagent and grams of $H_{2}$ gas produced. ( $M_{Z n} = 65$ , $M_{H C l} = 36.5$ , $M_{H_{2}} = 2$ )

Recall Solution 2.2

Moles: $n_{Z n} = \frac{13}{65} = 0.20$ mol, $n_{H C l} = \frac{14.6}{36.5} = 0.40$ mol. Ratios: $R_{Z n} = \frac{0.20}{1} = 0.20, R_{H C l} = \frac{0.40}{2} = 0.20.$ Equal again — an exact match, both fully consumed. Neither is in excess. Product $H_{2}$ : coefficient ratio $H_{2} : Z n = 1 : 1$ , so $n_{H_{2}} = 0.20$ mol $\Rightarrow m = 0.20 \times 2 = 0.40 g H_{2}$ .

Level 3 — Analysis

Goal: the "bigger mass fools you" cases plus leftover excess.

Exercise 3.1 (L3)

$4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ $27 g$ Al reacts with $48 g$ $O_{2}$ . Find the limiting reagent, the mass of $A l_{2} O_{3}$ formed, and the mass of leftover excess reactant. ( $M_{A l} = 27$ , $M_{O_{2}} = 32$ , $M_{A l_{2} O_{3}} = 102$ )

Recall Solution 3.1

Moles: $n_{A l} = \frac{27}{27} = 1.00$ mol, $n_{O_{2}} = \frac{48}{32} = 1.50$ mol. Ratios: $R_{A l} = \frac{1.00}{4} = 0.250, R_{O_{2}} = \frac{1.50}{3} = 0.500.$ $0.250 < 0.500 \Rightarrow$ Al is limiting (even though $O_{2}$ has more mass and more moles). Product: $n_{A l_{2} O_{3}} = n_{A l} \times \frac{2}{4} = 1.00 \times 0.5 = 0.500$ mol $\Rightarrow m = 0.500 \times 102 = 51.0 g$ . Leftover $O_{2}$ : $O_{2}$ used $= n_{A l} \times \frac{3}{4} = 1.00 \times 0.75 = 0.750$ mol. Leftover $= 1.50 - 0.750 = 0.750$ mol $\Rightarrow m = 0.750 \times 32 = 24.0 g O_{2}$ .

Exercise 3.2 (L3)

$2 N a + C l_{2} \to 2 N a C l$ $46 g$ Na reacts with $56.8 g$ $C l_{2}$ . Limiting reagent and mass of NaCl? ( $M_{N a} = 23$ , $M_{C l_{2}} = 71$ , $M_{N a C l} = 58.5$ )

Recall Solution 3.2

Moles: $n_{N a} = \frac{46}{23} = 2.00$ mol, $n_{C l_{2}} = \frac{56.8}{71} = 0.800$ mol. Ratios: $R_{N a} = \frac{2.00}{2} = 1.00, R_{C l_{2}} = \frac{0.800}{1} = 0.800.$ $0.800 < 1.00 \Rightarrow$ == $C l_{2}$ is limiting==. Product: $N a C l : C l_{2} = 2 : 1$ , so $n_{N a C l} = 0.800 \times 2 = 1.60$ mol $\Rightarrow m = 1.60 \times 58.5 = 93.6 g$ .

Level 4 — Synthesis

Goal: chain limiting reagent with percent yield, purity, and molarity.

Exercise 4.1 (L4)

$N_{2} + 3 H_{2} \to 2 N H_{3}$ $14 g$ $N_{2}$ reacts with $6 g$ $H_{2}$ . The lab actually collects $12.0 g$ $N H_{3}$ . Find the limiting reagent, the theoretical yield of $N H_{3}$ , and the percent yield. ( $M_{N_{2}} = 28$ , $M_{H_{2}} = 2$ , $M_{N H_{3}} = 17$ )

Recall Solution 4.1

Moles: $n_{N_{2}} = \frac{14}{28} = 0.500$ mol, $n_{H_{2}} = \frac{6}{2} = 3.00$ mol. Ratios: $R_{N_{2}} = \frac{0.500}{1} = 0.500$ , $R_{H_{2}} = \frac{3.00}{3} = 1.00$ . $0.500 < 1.00 \Rightarrow$ == $N_{2}$ is limiting==. Theoretical yield: $n_{N H_{3}} = 0.500 \times \frac{2}{1} = 1.00$ mol $\Rightarrow m_{theo} = 1.00 \times 17 = 17.0 g$ . Percent yield: $% yield = \frac{actual}{theoretical} \times 100% = \frac{12.0}{17.0} \times 100% = 70.6% .$

Exercise 4.2 (L4)

$C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$ $20.0 g$ of impure limestone that is $80.0%$ $C a C O_{3}$ by mass reacts with $500 mL$ of $1.00 mol/L$ HCl (see Concentration and molarity). Find the limiting reagent and the mass of $C O_{2}$ released. ( $M_{C a C O_{3}} = 100$ , $M_{C O_{2}} = 44$ )

Recall Solution 4.2

Pure $C a C O_{3}$ : $20.0 \times 0.800 = 16.0 g \Rightarrow n_{C a C O_{3}} = \frac{16.0}{100} = 0.160$ mol. Moles of HCl: $n = C \times V = 1.00 \frac{mol}{L} \times 0.500 L = 0.500$ mol. Ratios: $R_{C a C O_{3}} = \frac{0.160}{1} = 0.160, R_{H C l} = \frac{0.500}{2} = 0.250.$ $0.160 < 0.250 \Rightarrow$ == $C a C O_{3}$ is limiting==. Product $C O_{2}$ : $C O_{2} : C a C O_{3} = 1 : 1$ , so $n_{C O_{2}} = 0.160$ mol $\Rightarrow m = 0.160 \times 44 = 7.04 g C O_{2}$ .

Level 5 — Mastery

Goal: reverse problems, two-limiting-check consistency, and back-solving an unknown.

Exercise 5.1 (L5)

$2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ $30.0 g$ ethane ( $C_{2} H_{6}$ ) is burned in $112 g$ $O_{2}$ . Find the limiting reagent, the mass of $C O_{2}$ produced, and the mass of the excess reactant left over. ( $M_{C_{2} H_{6}} = 30$ , $M_{O_{2}} = 32$ , $M_{C O_{2}} = 44$ )

Recall Solution 5.1

Moles: $n_{C_{2} H_{6}} = \frac{30.0}{30} = 1.00$ mol, $n_{O_{2}} = \frac{112}{32} = 3.50$ mol. Ratios: $R_{C_{2} H_{6}} = \frac{1.00}{2} = 0.500, R_{O_{2}} = \frac{3.50}{7} = 0.500.$ Exactly equal — this is a stoichiometrically perfect mixture; both are fully consumed, no excess remains ( $0 g$ leftover). Product $C O_{2}$ : $C O_{2} : C_{2} H_{6} = 4 : 2 = 2$ , so $n_{C O_{2}} = 1.00 \times 2 = 2.00$ mol $\Rightarrow m = 2.00 \times 44 = 88.0 g$ .

Exercise 5.2 (L5)

$2 A g N O_{3} + B a C l_{2} \to 2 A g C l + B a (N O_{3})_{2}$ A solution containing $0.400$ mol $A g N O_{3}$ is mixed with $0.150$ mol $B a C l_{2}$ . After reaction, how many moles of each species (both products and any leftover reactant) are present?

Recall Solution 5.2

Ratios: $R_{A g N O_{3}} = \frac{0.400}{2} = 0.200$ , $R_{B a C l_{2}} = \frac{0.150}{1} = 0.150$ . $0.150 < 0.200 \Rightarrow$ == $B a C l_{2}$ is limiting==; the reaction runs $0.150$ full sets. $A g N O_{3}$ consumed: $0.150 \times 2 = 0.300$ mol $\Rightarrow$ leftover $= 0.400 - 0.300 = 0.100 mol A g N O_{3}$ . $A g C l$ formed: $A g C l : B a C l_{2} = 2 : 1 \Rightarrow 0.150 \times 2 = 0.300 mol$ . $B a (N O_{3})_{2}$ formed: $1 : 1$ with $B a C l_{2} \Rightarrow 0.150 mol$ . $B a C l_{2}$ left: $0$ (it was the limiting reagent — fully gone).

Exercise 5.3 (L5)

$3 H_{2} + N_{2} \to 2 N H_{3}$ You want $N H_{3}$ made with $H_{2}$ as the limiting reagent and exactly enough $N_{2}$ so that $0.30$ mol $N_{2}$ remains as excess afterwards. If you start with $9.0$ mol $H_{2}$ , what is the minimum starting moles of $N_{2}$ ?

Recall Solution 5.3

Reaction sets funded by $H_{2}$ : $R_{H_{2}} = \frac{9.0}{3} = 3.0$ sets (since $H_{2}$ is limiting). $N_{2}$ consumed: each set uses $1$ mol $N_{2} \Rightarrow 3.0 \times 1 = 3.0$ mol consumed. We also want $0.30$ mol $N_{2}$ left over, so $n_{N_{2}}^{start} = consumed + leftover = 3.0 + 0.30 = 3.3 mol N_{2} .$ Check $H_{2}$ is indeed limiting: $R_{N_{2}} = 3.3/1 = 3.3 > R_{H_{2}} = 3.0$ ✓.

Recall One-glance answer key

1.1 $O_{2}$ · 1.2 $H_{2}$ limiting, $2$ mol $N_{2}$ left · 2.1 $88$ g $C O_{2}$ (perfect mix) · 2.2 exact match, $0.40$ g $H_{2}$ · 3.1 Al limiting, $51.0$ g $A l_{2} O_{3}$ , $24.0$ g $O_{2}$ left · 3.2 $C l_{2}$ limiting, $93.6$ g NaCl · 4.1 $N_{2}$ limiting, $17.0$ g theoretical, $70.6%$ · 4.2 $C a C O_{3}$ limiting, $7.04$ g $C O_{2}$ · 5.1 perfect mix, $88.0$ g $C O_{2}$ , $0$ g excess · 5.2 $0.100$ mol $A g N O_{3}$ left, $0.300$ mol $A g C l$ , $0.150$ mol $B a (N O_{3})_{2}$ , $0$ $B a C l_{2}$ · 5.3 $3.3$ mol $N_{2}$ .

Connections

Limiting reagent problems — the parent method this page drills.
Balancing chemical equations — the coefficients $ν_{i}$ every ratio uses.
Mole concept and Avogadro number — why we compare moles, never grams.
Stoichiometric calculations — scaling from limiting reagent to any product.
Percent yield and purity — Exercises 4.1 and 4.2.
Concentration and molarity — Exercise 4.2's $n = C \times V$ step.
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