{"id":20680,"date":"2026-01-05T09:32:38","date_gmt":"2026-01-05T01:32:38","guid":{"rendered":"https:\/\/3d-universal.com\/en\/?p=20680"},"modified":"2026-01-05T09:39:55","modified_gmt":"2026-01-05T01:39:55","slug":"stoichiometry-and-chemical-reactions","status":"publish","type":"post","link":"https:\/\/3d-universal.com\/en\/blogs\/stoichiometry-and-chemical-reactions.html","title":{"rendered":"Stoichiometry and Chemical Reactions: NMAT Chemistry Review"},"content":{"rendered":"<p><!--more--><\/p>\n<h1>Stoichiometry and Chemical Reactions: NMAT Chemistry Review<\/h1>\n<p>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).<\/p>\n<p>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.<\/p>\n<h2>What Is Stoichiometry?<\/h2>\n<p>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.<\/p>\n<p>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.<\/p>\n<h2>Balancing Chemical Equations<\/h2>\n<p>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 \u201callowed\u201d numbers to use when relating reactants to products.<\/p>\n<p>Example (unbalanced):<\/p>\n<p>H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 H<sub>2<\/sub>O<\/p>\n<p>Balanced:<\/p>\n<p>2H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O<\/p>\n<p>This tells you: 2 moles of H<sub>2<\/sub> react with 1 mole of O<sub>2<\/sub> to produce 2 moles of H<sub>2<\/sub>O. If you use the unbalanced equation, every downstream number becomes wrong. In exam conditions, checking balance first prevents the most common stoichiometry errors.<\/p>\n<h2>The Mole Concept and Core Conversions<\/h2>\n<p>The mole is the bridge between microscopic particles and measurable quantities like mass and volume. One mole contains 6.022 \u00d7 10<sup>23<\/sup> entities (Avogadro\u2019s number). In stoichiometry, the mole is the universal \u201ccurrency.\u201d If a problem gives you grams, liters, or number of molecules, your first job is to convert to moles.<\/p>\n<p><strong>Key conversions:<\/strong><\/p>\n<ul>\n<li><strong>Mass to moles:<\/strong> moles = grams \u00f7 molar mass (g\/mol)<\/li>\n<li><strong>Moles to mass:<\/strong> grams = moles \u00d7 molar mass (g\/mol)<\/li>\n<li><strong>Moles to particles:<\/strong> particles = moles \u00d7 6.022 \u00d7 10<sup>23<\/sup><\/li>\n<li><strong>Particles to moles:<\/strong> moles = particles \u00f7 (6.022 \u00d7 10<sup>23<\/sup>)<\/li>\n<\/ul>\n<p>Molar mass is computed from the periodic table by summing the atomic masses according to the formula. For example, CO<sub>2<\/sub> has molar mass 12 + 2(16) = 44 g\/mol. CaCO<sub>3<\/sub> has molar mass 40 + 12 + 3(16) = 100 g\/mol. In NMAT calculations, approximate atomic masses are typically sufficient unless the problem demands precision.<\/p>\n<h2>How to Solve Stoichiometry Problems (NMAT Method)<\/h2>\n<p>Use a consistent workflow to avoid confusion:<\/p>\n<ol>\n<li><strong>Balance the equation.<\/strong><\/li>\n<li><strong>Convert the given quantity to moles.<\/strong> (If it\u2019s already in moles, move on.)<\/li>\n<li><strong>Use the mole ratio<\/strong> from coefficients to convert moles of given substance to moles of desired substance.<\/li>\n<li><strong>Convert moles to the requested unit<\/strong> (grams, liters at STP, particles, molarity-based volume, etc.).<\/li>\n<\/ol>\n<p><strong>Example:<\/strong> How many grams of CO<sub>2<\/sub> are produced when 25.0 g of CaCO<sub>3<\/sub> decomposes?<\/p>\n<p>Reaction: CaCO<sub>3<\/sub> \u2192 CaO + CO<sub>2<\/sub><\/p>\n<ul>\n<li>Balanced: already balanced (1:1:1)<\/li>\n<li>Moles CaCO<sub>3<\/sub> = 25.0 g \u00f7 100 g\/mol = 0.250 mol<\/li>\n<li>Mole ratio CaCO<sub>3<\/sub> : CO<sub>2<\/sub> = 1 : 1 \u2192 moles CO<sub>2<\/sub> = 0.250 mol<\/li>\n<li>Mass CO<sub>2<\/sub> = 0.250 mol \u00d7 44 g\/mol = 11.0 g<\/li>\n<\/ul>\n<p>This structure is highly testable and protects you from errors caused by skipping steps.<\/p>\n<h2>Reaction Types and Predicting Products<\/h2>\n<p>Recognizing common reaction patterns helps you predict products quickly, which is valuable in NMAT time pressure.<\/p>\n<ul>\n<li><strong>Synthesis (Combination):<\/strong> A + B \u2192 AB<\/li>\n<li><strong>Decomposition:<\/strong> AB \u2192 A + B<\/li>\n<li><strong>Single Replacement:<\/strong> A + BC \u2192 AC + B<\/li>\n<li><strong>Double Replacement:<\/strong> AB + CD \u2192 AD + CB<\/li>\n<li><strong>Combustion:<\/strong> hydrocarbon + O<sub>2<\/sub> \u2192 CO<sub>2<\/sub> + H<sub>2<\/sub>O (for complete combustion)<\/li>\n<\/ul>\n<p>Combustion is especially common: if a compound contains only carbon and hydrogen (and possibly oxygen), complete combustion typically yields CO<sub>2<\/sub> and H<sub>2<\/sub>O. Once products are known, you balance and proceed with stoichiometry.<\/p>\n<h2>Limiting Reactant and Excess Reactant<\/h2>\n<p>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.<\/p>\n<p><strong>NMAT-safe method:<\/strong><\/p>\n<ol>\n<li>Convert each reactant to moles.<\/li>\n<li>Use the balanced equation to calculate how much product each reactant could produce.<\/li>\n<li>The reactant that produces the smaller amount of product is the limiting reactant.<\/li>\n<\/ol>\n<p><strong>Example:<\/strong> For 2H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O, suppose you have 5.0 mol H<sub>2<\/sub> and 2.0 mol O<sub>2<\/sub>.<\/p>\n<ul>\n<li>If O<sub>2<\/sub> is limiting: 2.0 mol O<sub>2<\/sub> produces 4.0 mol H<sub>2<\/sub>O (because 1 mol O<sub>2<\/sub> \u2192 2 mol H<sub>2<\/sub>O).<\/li>\n<li>If H<sub>2<\/sub> is limiting: 5.0 mol H<sub>2<\/sub> produces 5.0 mol H<sub>2<\/sub>O (because 1 mol H<sub>2<\/sub> \u2192 1 mol H<sub>2<\/sub>O).<\/li>\n<li>The smaller product amount is 4.0 mol, so O<sub>2<\/sub> is limiting.<\/li>\n<\/ul>\n<p>To find excess remaining, compute how much of the excess reactant is consumed based on the limiting reactant, then subtract from the initial amount.<\/p>\n<h2>Theoretical Yield, Actual Yield, and Percent Yield<\/h2>\n<p>Stoichiometry gives a <strong>theoretical yield<\/strong>, the maximum product predicted by the balanced equation (and the limiting reactant if applicable). In real experiments, you often obtain an <strong>actual yield<\/strong> that is lower due to side reactions, incomplete reaction, or losses during purification.<\/p>\n<p><strong>Percent yield:<\/strong><\/p>\n<p>Percent yield = (actual yield \u00f7 theoretical yield) \u00d7 100%<\/p>\n<p><strong>Example:<\/strong> If theoretical yield is 20.0 g and actual yield is 15.0 g:<\/p>\n<p>Percent yield = (15.0 \u00f7 20.0) \u00d7 100% = 75%<\/p>\n<p>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.<\/p>\n<h2>Gas Stoichiometry and STP Shortcuts<\/h2>\n<p>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 \u201cstandard conditions.\u201d When STP is assumed, you can use:<\/p>\n<ul>\n<li>Volume (L) = moles \u00d7 22.4 L\/mol<\/li>\n<li>Moles = volume (L) \u00f7 22.4 L\/mol<\/li>\n<\/ul>\n<p><strong>Example:<\/strong> In CaCO<sub>3<\/sub> \u2192 CaO + CO<sub>2<\/sub>, 1.5 mol CaCO<sub>3<\/sub> produces 1.5 mol CO<sub>2<\/sub>. At STP, volume CO<sub>2<\/sub> = 1.5 \u00d7 22.4 = 33.6 L.<\/p>\n<p>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.<\/p>\n<h2>Solution Stoichiometry (Molarity and Reaction Amounts)<\/h2>\n<p>Solution problems frequently involve molarity, defined as moles of solute per liter of solution:<\/p>\n<p>M = moles \u00f7 liters<\/p>\n<p>From this, you can compute moles in a volume of solution:<\/p>\n<p>moles = M \u00d7 liters<\/p>\n<p><strong>Example:<\/strong> How many moles of NaCl are in 250 mL of 0.40 M NaCl?<\/p>\n<ul>\n<li>Convert volume: 250 mL = 0.250 L<\/li>\n<li>moles = 0.40 \u00d7 0.250 = 0.100 mol<\/li>\n<\/ul>\n<p>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.<\/p>\n<h2>Redox Stoichiometry (High-Yield Concepts)<\/h2>\n<p>Redox reactions involve electron transfer:<\/p>\n<ul>\n<li><strong>Oxidation:<\/strong> loss of electrons (increase in oxidation state)<\/li>\n<li><strong>Reduction:<\/strong> gain of electrons (decrease in oxidation state)<\/li>\n<\/ul>\n<p>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.<\/p>\n<h2>Common NMAT Mistakes and How to Avoid Them<\/h2>\n<ul>\n<li><strong>Using an unbalanced equation:<\/strong> always balance first.<\/li>\n<li><strong>Skipping mole conversion:<\/strong> convert grams, liters, and particles to moles before using ratios.<\/li>\n<li><strong>Mixing mass ratios with mole ratios:<\/strong> coefficients are mole ratios, not gram ratios.<\/li>\n<li><strong>Wrong limiting reactant assumption:<\/strong> calculate product from each reactant to identify the limiter.<\/li>\n<li><strong>Unit errors:<\/strong> keep track of g, mol, L, and particles at every step.<\/li>\n<\/ul>\n<p>A practical exam habit is to write the unit at each line of calculation. If units cancel properly, your setup is likely correct.<\/p>\n<h2>NMAT Exam Strategy for Stoichiometry Questions<\/h2>\n<p>To perform well under time pressure, combine conceptual clarity with consistent setup:<\/p>\n<ul>\n<li>Write the balanced equation immediately, even if you think it is already balanced.<\/li>\n<li>Convert all given quantities to moles before doing anything else.<\/li>\n<li>Use coefficients as your only stoichiometric ratios.<\/li>\n<li>Estimate to catch unreasonable answers (e.g., negative masses, volumes too large for given moles).<\/li>\n<li>Practice timed drills: speed improves dramatically when your setup becomes automatic.<\/li>\n<\/ul>\n<p>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.<\/p>\n<h2>Summary<\/h2>\n<p>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\u2014balance, convert to moles, apply ratios, convert units\u2014you can solve most stoichiometry problems efficiently. With regular practice, these problems become predictable and highly scorable.<\/p>\n<h2>Problem Sets<\/h2>\n<h3>Set 1: Balancing Equations and Reaction Types<\/h3>\n<ol>\n<li>Balance the equation: Al + O<sub>2<\/sub> \u2192 Al<sub>2<\/sub>O<sub>3<\/sub><\/li>\n<li>Balance the combustion equation: C<sub>3<\/sub>H<sub>8<\/sub> + O<sub>2<\/sub> \u2192 CO<sub>2<\/sub> + H<sub>2<\/sub>O<\/li>\n<li>Identify the reaction type (synthesis, decomposition, single replacement, double replacement, combustion): 2KClO<sub>3<\/sub> \u2192 2KCl + 3O<sub>2<\/sub><\/li>\n<li>Identify the reaction type: Zn + 2HCl \u2192 ZnCl<sub>2<\/sub> + H<sub>2<\/sub><\/li>\n<li>Balance the equation: Fe + O<sub>2<\/sub> \u2192 Fe<sub>2<\/sub>O<sub>3<\/sub><\/li>\n<\/ol>\n<h3>Set 2: Mole\u2013Mass and Particle Conversions<\/h3>\n<ol start=\"6\">\n<li>How many moles are in 36.0 g of H<sub>2<\/sub>O?<\/li>\n<li>What mass (g) is 0.50 mol of CO<sub>2<\/sub>?<\/li>\n<li>How many molecules are in 0.25 mol of NH<sub>3<\/sub>?<\/li>\n<li>How many grams of NaCl are in 2.00 mol of NaCl?<\/li>\n<li>Find the molar mass of Ca(OH)<sub>2<\/sub>.<\/li>\n<\/ol>\n<h3>Set 3: Basic Stoichiometry (Mole Ratios)<\/h3>\n<ol start=\"11\">\n<li>For 2H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O, how many moles of H<sub>2<\/sub>O form from 3.0 mol H<sub>2<\/sub> (excess O<sub>2<\/sub>)?<\/li>\n<li>For N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub>, how many moles of NH<sub>3<\/sub> form from 2.0 mol N<sub>2<\/sub> (excess H<sub>2<\/sub>)?<\/li>\n<li>For 2Na + Cl<sub>2<\/sub> \u2192 2NaCl, how many moles of Cl<sub>2<\/sub> are required to produce 5.0 mol NaCl?<\/li>\n<li>For CaCO<sub>3<\/sub> \u2192 CaO + CO<sub>2<\/sub>, how many grams of CO<sub>2<\/sub> form from 25.0 g CaCO<sub>3<\/sub>?<\/li>\n<li>For 4Fe + 3O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub>, how many moles of O<sub>2<\/sub> are needed to react with 6.0 mol Fe?<\/li>\n<\/ol>\n<h3>Set 4: Limiting Reactant<\/h3>\n<ol start=\"16\">\n<li>For 2H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O, if you have 5.0 mol H<sub>2<\/sub> and 2.0 mol O<sub>2<\/sub>, identify the limiting reactant and calculate moles of H<sub>2<\/sub>O produced.<\/li>\n<li>For N<sub>2<\/sub> + 3H<sub>2<\/sub> \u2192 2NH<sub>3<\/sub>, if you have 1.0 mol N<sub>2<\/sub> and 2.0 mol H<sub>2<\/sub>, identify the limiting reactant and calculate moles of NH<sub>3<\/sub> produced.<\/li>\n<li>For 2Na + Cl<sub>2<\/sub> \u2192 2NaCl, if you have 3.0 mol Na and 1.0 mol Cl<sub>2<\/sub>, identify the limiting reactant and calculate moles of NaCl produced.<\/li>\n<li>For CH<sub>4<\/sub> + 2O<sub>2<\/sub> \u2192 CO<sub>2<\/sub> + 2H<sub>2<\/sub>O, if you have 2.0 mol CH<sub>4<\/sub> and 3.0 mol O<sub>2<\/sub>, identify the limiting reactant and calculate moles of CO<sub>2<\/sub> produced.<\/li>\n<li>For CaO + CO<sub>2<\/sub> \u2192 CaCO<sub>3<\/sub>, if you have 0.80 mol CaO and 1.20 mol CO<sub>2<\/sub>, identify the limiting reactant and calculate moles of CaCO<sub>3<\/sub> produced.<\/li>\n<\/ol>\n<h3>Set 5: Percent Yield<\/h3>\n<ol start=\"21\">\n<li>For 2H<sub>2<\/sub> + O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O, if the theoretical yield is 18.0 g and the actual yield is 15.3 g, calculate percent yield.<\/li>\n<li>A reaction has a theoretical yield of 40.0 g. The actual yield is 30.0 g. Find the percent yield.<\/li>\n<li>If percent yield is 80% and the theoretical yield is 50.0 g, what is the actual yield?<\/li>\n<li>If actual yield is 12.0 g and percent yield is 60%, what is the theoretical yield?<\/li>\n<li>A student expects 10.0 g of product but obtains 8.50 g. Compute percent yield.<\/li>\n<\/ol>\n<h3>Set 6: Gas Stoichiometry at STP<\/h3>\n<p>(Use 22.4 L\/mol at STP.)<\/p>\n<ol start=\"26\">\n<li>How many liters of O<sub>2<\/sub> at STP are produced from 2.0 mol KClO<sub>3<\/sub> in 2KClO<sub>3<\/sub> \u2192 2KCl + 3O<sub>2<\/sub>?<\/li>\n<li>How many liters of CO<sub>2<\/sub> at STP are produced from 1.5 mol CaCO<sub>3<\/sub> in CaCO<sub>3<\/sub> \u2192 CaO + CO<sub>2<\/sub>?<\/li>\n<li>How many moles of O<sub>2<\/sub> are in 11.2 L of O<sub>2<\/sub> at STP?<\/li>\n<li>How many liters of H<sub>2<\/sub> at STP are produced when 4.0 mol Zn reacts in Zn + 2HCl \u2192 ZnCl<sub>2<\/sub> + H<sub>2<\/sub>?<\/li>\n<li>How many moles of CO<sub>2<\/sub> correspond to 44.8 L of CO<sub>2<\/sub> at STP?<\/li>\n<\/ol>\n<h3>Set 7: Solution Stoichiometry (Molarity)<\/h3>\n<ol start=\"31\">\n<li>How many moles of NaCl are in 250 mL of 0.40 M NaCl?<\/li>\n<li>What is the molarity of a solution containing 0.75 mol solute in 0.50 L solution?<\/li>\n<li>How many grams of glucose (C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>, molar mass = 180 g\/mol) are in 0.20 mol?<\/li>\n<li>How many moles of HCl are present in 100 mL of 2.0 M HCl?<\/li>\n<li>If 0.50 L of 0.10 M NaOH reacts completely, how many moles of NaOH reacted?<\/li>\n<\/ol>\n<h2>Answer Keys<\/h2>\n<h3>Set 1 Answers<\/h3>\n<ol>\n<li>4Al + 3O<sub>2<\/sub> \u2192 2Al<sub>2<\/sub>O<sub>3<\/sub><\/li>\n<li>C<sub>3<\/sub>H<sub>8<\/sub> + 5O<sub>2<\/sub> \u2192 3CO<sub>2<\/sub> + 4H<sub>2<\/sub>O<\/li>\n<li>Decomposition<\/li>\n<li>Single replacement<\/li>\n<li>4Fe + 3O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub><\/li>\n<\/ol>\n<h3>Set 2 Answers<\/h3>\n<ol start=\"6\">\n<li>36.0 g \u00f7 18.0 g\/mol = 2.00 mol<\/li>\n<li>0.50 mol \u00d7 44.0 g\/mol = 22.0 g<\/li>\n<li>0.25 mol \u00d7 6.022 \u00d7 10<sup>23<\/sup> = 1.51 \u00d7 10<sup>23<\/sup> molecules<\/li>\n<li>2.00 mol \u00d7 58.5 g\/mol = 117 g<\/li>\n<li>Ca(OH)<sub>2<\/sub> = 40 + 2(16 + 1) = 74 g\/mol<\/li>\n<\/ol>\n<h3>Set 3 Answers<\/h3>\n<ol start=\"11\">\n<li>2H<sub>2<\/sub> : 2H<sub>2<\/sub>O = 1:1, so 3.0 mol H<sub>2<\/sub> \u2192 3.0 mol H<sub>2<\/sub>O<\/li>\n<li>1 mol N<sub>2<\/sub> \u2192 2 mol NH<sub>3<\/sub>, so 2.0 mol N<sub>2<\/sub> \u2192 4.0 mol NH<sub>3<\/sub><\/li>\n<li>Cl<sub>2<\/sub> : NaCl = 1:2, so 5.0 mol NaCl requires 2.5 mol Cl<sub>2<\/sub><\/li>\n<li>25.0 g \u00f7 100 g\/mol = 0.250 mol CaCO<sub>3<\/sub> \u2192 0.250 mol CO<sub>2<\/sub>; mass = 0.250 \u00d7 44 = 11.0 g<\/li>\n<li>Fe:O<sub>2<\/sub> = 4:3, so 6.0 mol Fe requires 6.0 \u00d7 (3\/4) = 4.5 mol O<sub>2<\/sub><\/li>\n<\/ol>\n<h3>Set 4 Answers<\/h3>\n<ol start=\"16\">\n<li>O<sub>2<\/sub> is limiting. 2.0 mol O<sub>2<\/sub> \u2192 4.0 mol H<sub>2<\/sub>O<\/li>\n<li>H<sub>2<\/sub> is limiting. 2.0 mol H<sub>2<\/sub> \u00d7 (2\/3) = 1.33 mol NH<sub>3<\/sub><\/li>\n<li>Cl<sub>2<\/sub> is limiting. 1.0 mol Cl<sub>2<\/sub> \u2192 2.0 mol NaCl<\/li>\n<li>O<sub>2<\/sub> is limiting. 3.0 mol O<sub>2<\/sub> \u00d7 (1\/2) = 1.5 mol CO<sub>2<\/sub><\/li>\n<li>CaO is limiting. 0.80 mol CaO \u2192 0.80 mol CaCO<sub>3<\/sub><\/li>\n<\/ol>\n<h3>Set 5 Answers<\/h3>\n<ol start=\"21\">\n<li>(15.3 \u00f7 18.0) \u00d7 100 = 85.0%<\/li>\n<li>(30.0 \u00f7 40.0) \u00d7 100 = 75.0%<\/li>\n<li>Actual yield = 0.80 \u00d7 50.0 g = 40.0 g<\/li>\n<li>Theoretical yield = 12.0 g \u00f7 0.60 = 20.0 g<\/li>\n<li>(8.50 \u00f7 10.0) \u00d7 100 = 85.0%<\/li>\n<\/ol>\n<h3>Set 6 Answers<\/h3>\n<ol start=\"26\">\n<li>2.0 mol KClO<sub>3<\/sub> \u2192 3.0 mol O<sub>2<\/sub>; volume = 3.0 \u00d7 22.4 = 67.2 L<\/li>\n<li>1.5 mol CaCO<sub>3<\/sub> \u2192 1.5 mol CO<sub>2<\/sub>; volume = 1.5 \u00d7 22.4 = 33.6 L<\/li>\n<li>11.2 \u00f7 22.4 = 0.50 mol<\/li>\n<li>4.0 mol Zn \u2192 4.0 mol H<sub>2<\/sub>; volume = 4.0 \u00d7 22.4 = 89.6 L<\/li>\n<li>44.8 \u00f7 22.4 = 2.00 mol<\/li>\n<\/ol>\n<h3>Set 7 Answers<\/h3>\n<ol start=\"31\">\n<li>250 mL = 0.250 L; moles = 0.40 \u00d7 0.250 = 0.100 mol<\/li>\n<li>M = 0.75 \u00f7 0.50 = 1.5 M<\/li>\n<li>Mass = 0.20 \u00d7 180 = 36 g<\/li>\n<li>100 mL = 0.100 L; moles = 2.0 \u00d7 0.100 = 0.200 mol<\/li>\n<li>Moles = 0.10 \u00d7 0.50 = 0.050 mol<\/li>\n<\/ol>\n<blockquote class=\"wp-embedded-content\" data-secret=\"ms4XDSBxeb\"><p><a href=\"https:\/\/3d-universal.com\/en\/blogs\/nmat-chemistry-review.html\">NMAT Chemistry Review: NMAT Study Guide<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; visibility: hidden;\" title=\"&#8220;NMAT Chemistry Review: NMAT Study Guide&#8221; &#8212; Study English at 3D ACADEMY, a Language School in Cebu, Philippines\" src=\"https:\/\/3d-universal.com\/en\/blogs\/nmat-chemistry-review.html\/embed#?secret=rbW3EBmjmg#?secret=ms4XDSBxeb\" data-secret=\"ms4XDSBxeb\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":20683,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_kad_blocks_custom_css":"","_kad_blocks_head_custom_js":"","_kad_blocks_body_custom_js":"","_kad_blocks_footer_custom_js":"","footnotes":""},"categories":[116],"tags":[],"class_list":["post-20680","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nmat-chemistry-review"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v25.6 (Yoast SEO v25.6) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Stoichiometry and Chemical Reactions: NMAT Chemistry Review - Study English at 3D ACADEMY, a Language School in Cebu, Philippines<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" 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