Stoichiometry and Chemical Reactions: NMAT Chemistry Review

Stoichiometry and chemical reactions are core topics in NMAT Chemistry because they test both conceptual understanding and fast, accurate quantitative reasoning. Many NMAT items require you to translate words into balanced equations, interpret mole ratios, and move confidently between mass, moles, particles, gas volume, and solution concentration. If you can set up stoichiometry systematically, you can solve a wide range of problems: limiting reactants, percent yield, combustion analysis, gas stoichiometry, and solution-based reactions (including titration-style questions).

This review focuses on the most tested skills for NMAT: balancing equations, the mole concept, stoichiometric conversion pathways, limiting reactant logic, yield calculations, and recognizing reaction types. The goal is to build a repeatable method that you can apply under time pressure.

What Is Stoichiometry?

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It is grounded in the law of conservation of mass: atoms are neither created nor destroyed in ordinary chemical reactions. Because atoms must be conserved, a balanced chemical equation provides fixed mole ratios between substances. Stoichiometry uses those ratios to predict how much product forms from given reactants, or how much reactant is needed to obtain a target amount of product.

In NMAT-style questions, stoichiometry often appears as multi-step problems that combine concepts such as molar mass, solution molarity, gas volume at standard conditions, and sometimes redox or acid-base chemistry. The reliable approach is always the same: balance the equation, convert to moles, apply mole ratios, then convert to the asked unit.

Balancing Chemical Equations

Before you do any calculation, ensure the reaction is balanced. A balanced equation has the same number of each type of atom on both sides. The coefficients (numbers in front of formulas) represent mole ratios, and they are the only “allowed” numbers to use when relating reactants to products.

Example (unbalanced):

H₂ + O₂ → H₂O

Balanced:

2H₂ + O₂ → 2H₂O

This tells you: 2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O. If you use the unbalanced equation, every downstream number becomes wrong. In exam conditions, checking balance first prevents the most common stoichiometry errors.

The Mole Concept and Core Conversions

The mole is the bridge between microscopic particles and measurable quantities like mass and volume. One mole contains 6.022 × 10²³ entities (Avogadro’s number). In stoichiometry, the mole is the universal “currency.” If a problem gives you grams, liters, or number of molecules, your first job is to convert to moles.

Key conversions:

• Mass to moles: moles = grams ÷ molar mass (g/mol)
• Moles to mass: grams = moles × molar mass (g/mol)
• Moles to particles: particles = moles × 6.022 × 10²³
• Particles to moles: moles = particles ÷ (6.022 × 10²³)

Molar mass is computed from the periodic table by summing the atomic masses according to the formula. For example, CO₂ has molar mass 12 + 2(16) = 44 g/mol. CaCO₃ has molar mass 40 + 12 + 3(16) = 100 g/mol. In NMAT calculations, approximate atomic masses are typically sufficient unless the problem demands precision.

How to Solve Stoichiometry Problems (NMAT Method)

Use a consistent workflow to avoid confusion:

1. Balance the equation.
2. Convert the given quantity to moles. (If it’s already in moles, move on.)
3. Use the mole ratio from coefficients to convert moles of given substance to moles of desired substance.
4. Convert moles to the requested unit (grams, liters at STP, particles, molarity-based volume, etc.).

Example: How many grams of CO₂ are produced when 25.0 g of CaCO₃ decomposes?

Reaction: CaCO₃ → CaO + CO₂

• Balanced: already balanced (1:1:1)
• Moles CaCO₃ = 25.0 g ÷ 100 g/mol = 0.250 mol
• Mole ratio CaCO₃ : CO₂ = 1 : 1 → moles CO₂ = 0.250 mol
• Mass CO₂ = 0.250 mol × 44 g/mol = 11.0 g

This structure is highly testable and protects you from errors caused by skipping steps.

Reaction Types and Predicting Products

Recognizing common reaction patterns helps you predict products quickly, which is valuable in NMAT time pressure.

• Synthesis (Combination): A + B → AB
• Decomposition: AB → A + B
• Single Replacement: A + BC → AC + B
• Double Replacement: AB + CD → AD + CB
• Combustion: hydrocarbon + O₂ → CO₂ + H₂O (for complete combustion)

Combustion is especially common: if a compound contains only carbon and hydrogen (and possibly oxygen), complete combustion typically yields CO₂ and H₂O. Once products are known, you balance and proceed with stoichiometry.

Limiting Reactant and Excess Reactant

When more than one reactant is present, the reaction stops when the limiting reactant is fully consumed. The limiting reactant determines the maximum possible amount of product. The other reactant(s) are in excess and remain partly unreacted.

NMAT-safe method:

1. Convert each reactant to moles.
2. Use the balanced equation to calculate how much product each reactant could produce.
3. The reactant that produces the smaller amount of product is the limiting reactant.

Example: For 2H₂ + O₂ → 2H₂O, suppose you have 5.0 mol H₂ and 2.0 mol O₂.

• If O₂ is limiting: 2.0 mol O₂ produces 4.0 mol H₂O (because 1 mol O₂ → 2 mol H₂O).
• If H₂ is limiting: 5.0 mol H₂ produces 5.0 mol H₂O (because 1 mol H₂ → 1 mol H₂O).
• The smaller product amount is 4.0 mol, so O₂ is limiting.

To find excess remaining, compute how much of the excess reactant is consumed based on the limiting reactant, then subtract from the initial amount.

Theoretical Yield, Actual Yield, and Percent Yield

Stoichiometry gives a theoretical yield, the maximum product predicted by the balanced equation (and the limiting reactant if applicable). In real experiments, you often obtain an actual yield that is lower due to side reactions, incomplete reaction, or losses during purification.

Percent yield:

Percent yield = (actual yield ÷ theoretical yield) × 100%

Example: If theoretical yield is 20.0 g and actual yield is 15.0 g:

Percent yield = (15.0 ÷ 20.0) × 100% = 75%

NMAT questions may give percent yield and ask for actual yield, or give actual yield and ask for theoretical yield. Treat the formula as an algebraic relationship and rearrange carefully.

Gas Stoichiometry and STP Shortcuts

Gas stoichiometry links moles to gas volume. At standard temperature and pressure (STP), 1 mole of an ideal gas occupies 22.4 L. NMAT problems may explicitly state STP or imply “standard conditions.” When STP is assumed, you can use:

• Volume (L) = moles × 22.4 L/mol
• Moles = volume (L) ÷ 22.4 L/mol

Example: In CaCO₃ → CaO + CO₂, 1.5 mol CaCO₃ produces 1.5 mol CO₂. At STP, volume CO₂ = 1.5 × 22.4 = 33.6 L.

If conditions are not STP, the problem may require PV = nRT. In that case, still use stoichiometry to find moles first, then apply the gas law.

Solution Stoichiometry (Molarity and Reaction Amounts)

Solution problems frequently involve molarity, defined as moles of solute per liter of solution:

M = moles ÷ liters

From this, you can compute moles in a volume of solution:

moles = M × liters

Example: How many moles of NaCl are in 250 mL of 0.40 M NaCl?

• Convert volume: 250 mL = 0.250 L
• moles = 0.40 × 0.250 = 0.100 mol

In reaction stoichiometry involving solutions (such as acid-base neutralization), the key is to convert each solution volume to moles using molarity, then use the balanced equation to relate reactants. Even if the problem resembles a titration, the same mole-ratio logic applies.

Redox Stoichiometry (High-Yield Concepts)

Redox reactions involve electron transfer:

• Oxidation: loss of electrons (increase in oxidation state)
• Reduction: gain of electrons (decrease in oxidation state)

NMAT questions may ask you to identify oxidizing and reducing agents, or apply stoichiometry after a redox equation is balanced. The most common challenge is balancing correctly first. Once the equation is balanced, stoichiometry proceeds normally: convert to moles, apply ratios, and convert to desired units.

Common NMAT Mistakes and How to Avoid Them

• Using an unbalanced equation: always balance first.
• Skipping mole conversion: convert grams, liters, and particles to moles before using ratios.
• Mixing mass ratios with mole ratios: coefficients are mole ratios, not gram ratios.
• Wrong limiting reactant assumption: calculate product from each reactant to identify the limiter.
• Unit errors: keep track of g, mol, L, and particles at every step.

A practical exam habit is to write the unit at each line of calculation. If units cancel properly, your setup is likely correct.

NMAT Exam Strategy for Stoichiometry Questions

To perform well under time pressure, combine conceptual clarity with consistent setup:

• Write the balanced equation immediately, even if you think it is already balanced.
• Convert all given quantities to moles before doing anything else.
• Use coefficients as your only stoichiometric ratios.
• Estimate to catch unreasonable answers (e.g., negative masses, volumes too large for given moles).
• Practice timed drills: speed improves dramatically when your setup becomes automatic.

Stoichiometry rewards a disciplined method. If you train yourself to follow the same pathway every time, you will reduce careless mistakes and increase your score consistency on NMAT Chemistry.

Summary

Stoichiometry is the quantitative language of chemical reactions. For NMAT Chemistry, the highest-impact skills are balancing equations, converting to moles, applying mole ratios, identifying limiting reactants, calculating yields, and handling gases and solutions. If you master the standard workflow—balance, convert to moles, apply ratios, convert units—you can solve most stoichiometry problems efficiently. With regular practice, these problems become predictable and highly scorable.

Problem Sets

Set 1: Balancing Equations and Reaction Types

1. Balance the equation: Al + O₂ → Al₂O₃
2. Balance the combustion equation: C₃H₈ + O₂ → CO₂ + H₂O
3. Identify the reaction type (synthesis, decomposition, single replacement, double replacement, combustion): 2KClO₃ → 2KCl + 3O₂
4. Identify the reaction type: Zn + 2HCl → ZnCl₂ + H₂
5. Balance the equation: Fe + O₂ → Fe₂O₃

Set 2: Mole–Mass and Particle Conversions

6. How many moles are in 36.0 g of H₂O?
7. What mass (g) is 0.50 mol of CO₂?
8. How many molecules are in 0.25 mol of NH₃?
9. How many grams of NaCl are in 2.00 mol of NaCl?
10. Find the molar mass of Ca(OH)₂.

Set 3: Basic Stoichiometry (Mole Ratios)

11. For 2H₂ + O₂ → 2H₂O, how many moles of H₂O form from 3.0 mol H₂ (excess O₂)?
12. For N₂ + 3H₂ → 2NH₃, how many moles of NH₃ form from 2.0 mol N₂ (excess H₂)?
13. For 2Na + Cl₂ → 2NaCl, how many moles of Cl₂ are required to produce 5.0 mol NaCl?
14. For CaCO₃ → CaO + CO₂, how many grams of CO₂ form from 25.0 g CaCO₃?
15. For 4Fe + 3O₂ → 2Fe₂O₃, how many moles of O₂ are needed to react with 6.0 mol Fe?

Set 4: Limiting Reactant

16. For 2H₂ + O₂ → 2H₂O, if you have 5.0 mol H₂ and 2.0 mol O₂, identify the limiting reactant and calculate moles of H₂O produced.
17. For N₂ + 3H₂ → 2NH₃, if you have 1.0 mol N₂ and 2.0 mol H₂, identify the limiting reactant and calculate moles of NH₃ produced.
18. For 2Na + Cl₂ → 2NaCl, if you have 3.0 mol Na and 1.0 mol Cl₂, identify the limiting reactant and calculate moles of NaCl produced.
19. For CH₄ + 2O₂ → CO₂ + 2H₂O, if you have 2.0 mol CH₄ and 3.0 mol O₂, identify the limiting reactant and calculate moles of CO₂ produced.
20. For CaO + CO₂ → CaCO₃, if you have 0.80 mol CaO and 1.20 mol CO₂, identify the limiting reactant and calculate moles of CaCO₃ produced.

Set 5: Percent Yield

21. For 2H₂ + O₂ → 2H₂O, if the theoretical yield is 18.0 g and the actual yield is 15.3 g, calculate percent yield.
22. A reaction has a theoretical yield of 40.0 g. The actual yield is 30.0 g. Find the percent yield.
23. If percent yield is 80% and the theoretical yield is 50.0 g, what is the actual yield?
24. If actual yield is 12.0 g and percent yield is 60%, what is the theoretical yield?
25. A student expects 10.0 g of product but obtains 8.50 g. Compute percent yield.

Set 6: Gas Stoichiometry at STP

(Use 22.4 L/mol at STP.)

26. How many liters of O₂ at STP are produced from 2.0 mol KClO₃ in 2KClO₃ → 2KCl + 3O₂?
27. How many liters of CO₂ at STP are produced from 1.5 mol CaCO₃ in CaCO₃ → CaO + CO₂?
28. How many moles of O₂ are in 11.2 L of O₂ at STP?
29. How many liters of H₂ at STP are produced when 4.0 mol Zn reacts in Zn + 2HCl → ZnCl₂ + H₂?
30. How many moles of CO₂ correspond to 44.8 L of CO₂ at STP?

Set 7: Solution Stoichiometry (Molarity)

31. How many moles of NaCl are in 250 mL of 0.40 M NaCl?
32. What is the molarity of a solution containing 0.75 mol solute in 0.50 L solution?
33. How many grams of glucose (C₆H₁₂O₆, molar mass = 180 g/mol) are in 0.20 mol?
34. How many moles of HCl are present in 100 mL of 2.0 M HCl?
35. If 0.50 L of 0.10 M NaOH reacts completely, how many moles of NaOH reacted?

Answer Keys

Set 1 Answers

1. 4Al + 3O₂ → 2Al₂O₃
2. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
3. Decomposition
4. Single replacement
5. 4Fe + 3O₂ → 2Fe₂O₃

Set 2 Answers

6. 36.0 g ÷ 18.0 g/mol = 2.00 mol
7. 0.50 mol × 44.0 g/mol = 22.0 g
8. 0.25 mol × 6.022 × 10²³ = 1.51 × 10²³ molecules
9. 2.00 mol × 58.5 g/mol = 117 g
10. Ca(OH)₂ = 40 + 2(16 + 1) = 74 g/mol

Set 3 Answers

11. 2H₂ : 2H₂O = 1:1, so 3.0 mol H₂ → 3.0 mol H₂O
12. 1 mol N₂ → 2 mol NH₃, so 2.0 mol N₂ → 4.0 mol NH₃
13. Cl₂ : NaCl = 1:2, so 5.0 mol NaCl requires 2.5 mol Cl₂
14. 25.0 g ÷ 100 g/mol = 0.250 mol CaCO₃ → 0.250 mol CO₂; mass = 0.250 × 44 = 11.0 g
15. Fe:O₂ = 4:3, so 6.0 mol Fe requires 6.0 × (3/4) = 4.5 mol O₂

Set 4 Answers

16. O₂ is limiting. 2.0 mol O₂ → 4.0 mol H₂O
17. H₂ is limiting. 2.0 mol H₂ × (2/3) = 1.33 mol NH₃
18. Cl₂ is limiting. 1.0 mol Cl₂ → 2.0 mol NaCl
19. O₂ is limiting. 3.0 mol O₂ × (1/2) = 1.5 mol CO₂
20. CaO is limiting. 0.80 mol CaO → 0.80 mol CaCO₃

Set 5 Answers

21. (15.3 ÷ 18.0) × 100 = 85.0%
22. (30.0 ÷ 40.0) × 100 = 75.0%
23. Actual yield = 0.80 × 50.0 g = 40.0 g
24. Theoretical yield = 12.0 g ÷ 0.60 = 20.0 g
25. (8.50 ÷ 10.0) × 100 = 85.0%

Set 6 Answers

26. 2.0 mol KClO₃ → 3.0 mol O₂; volume = 3.0 × 22.4 = 67.2 L
27. 1.5 mol CaCO₃ → 1.5 mol CO₂; volume = 1.5 × 22.4 = 33.6 L
28. 11.2 ÷ 22.4 = 0.50 mol
29. 4.0 mol Zn → 4.0 mol H₂; volume = 4.0 × 22.4 = 89.6 L
30. 44.8 ÷ 22.4 = 2.00 mol

Set 7 Answers

31. 250 mL = 0.250 L; moles = 0.40 × 0.250 = 0.100 mol
32. M = 0.75 ÷ 0.50 = 1.5 M
33. Mass = 0.20 × 180 = 36 g
34. 100 mL = 0.100 L; moles = 2.0 × 0.100 = 0.200 mol
35. Moles = 0.10 × 0.50 = 0.050 mol

NMAT Chemistry Review: NMAT Study Guide