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Quizzes > High School Quizzes > Science

Stoich Review Worksheet: Practice Quiz

Boost your skills with interactive practice exercises

Difficulty: Moderate
Grade: Grade 11
Study OutcomesCheat Sheet
Colorful paper art promoting The Stoich Review Challenge, a high school chemistry quiz.

What is a mole in chemistry?
A unit measuring time
A measure of temperature
A unit representing 6.022 x 10^23 particles
A unit of length
A mole is a fundamental unit in chemistry used to quantify the amount of substance. It is defined as 6.022 x 10^23 particles, which is Avogadro's number.
Which of the following best describes stoichiometry?
The study of chemical bonding
The study of reaction mechanisms
The study of the quantitative relationships in chemical reactions
The study of periodic trends in elements
Stoichiometry focuses on calculating the quantitative relationships between reactants and products in a chemical reaction. This knowledge is essential for predicting yields and balancing equations.
What does a balanced chemical equation ensure?
It shows the conservation of atoms and mass
It shows that both sides have the same number of molecules
It indicates that volume and pressure are constant
It ensures that energy is produced
A balanced chemical equation confirms that the number of atoms for each element is the same on both sides of the equation, reflecting the conservation of mass. It does not necessarily imply that the number of molecules or energy values are identical.
What is the role of a coefficient in a balanced equation?
It indicates the mass of a substance
It indicates the concentration of a solution
It indicates the number of moles of a substance
It indicates the volume of a gas
Coefficients represent the number of moles of each substance participating in a chemical reaction. They are used to establish the mole ratios required for stoichiometric calculations.
Which constant is used to represent the number of particles in one mole?
6.022 x 10^23
3.00 x 10^8 m/s
9.81 m/s²
1.602 x 10^-19
Avogadro's number, 6.022 x 10^23, defines the number of particles in one mole of a substance. This constant is essential for converting between moles and the number of atoms or molecules.
How many moles are present in 36 grams of H2O (molar mass = 18 g/mol)?
3 moles
1 mole
4 moles
2 moles
To determine the number of moles, divide the mass of the substance by its molar mass. In this case, 36 g divided by 18 g/mol yields 2 moles.
What is the mass of 3 moles of CO2 (molar mass = 44 g/mol)?
44 grams
176 grams
132 grams
88 grams
Multiplying the number of moles by the molar mass gives the overall mass. For 3 moles of CO2, the calculation is 3 x 44 g/mol, which equals 132 grams.
If 5 moles of A react with 7 moles of B in a reaction with a 1:2 stoichiometric ratio (A:B), which reactant is limiting?
A is limiting
B is limiting
Neither, both are in excess
Both react equally limit the reaction
For every mole of A, 2 moles of B are needed. With 5 moles of A requiring 10 moles of B and only 7 moles available, B is the limiting reactant.
In the reaction 2H2 + O2 â†' 2H2O, if you have 4 moles of H2 and 3 moles of O2, which reactant is limiting?
Both are in excess
H2 is limiting
Neither is limiting
O2 is limiting
The reaction requires 2 moles of H2 for every 1 mole of O2. Here, 4 moles of H2 would need 2 moles of O2, so the excess oxygen confirms that H2 is the limiting reactant.
What is the percent yield if the theoretical yield is 50 grams and the actual yield is 40 grams?
90%
60%
70%
80%
Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. (40 g / 50 g) x 100 equals 80%.
What is the first step in determining an empirical formula from percent composition data?
Divide the percent composition of each element by its atomic mass
Multiply the percent composition by the atomic mass
Convert the percent composition directly into moles
Add the percent compositions together
The first step in finding an empirical formula is to convert the percentage of each element to moles by dividing by the atomic mass. This generates the mole ratios needed to determine the simplest whole-number ratio.
What is the molar mass of CaCO3?
80 g/mol
90 g/mol
120 g/mol
100 g/mol
Calcium carbonate (CaCO3) is composed of calcium (40 g/mol), carbon (12 g/mol), and three oxygen atoms (16 g/mol each, totaling 48 g/mol). Adding these gives a molar mass of 100 g/mol.
What does a coefficient in a balanced equation represent?
The volume of the substance
The number of moles of the substance
The mass of the substance
The energy of the reaction
Coefficients indicate the number of moles of each substance involved in a chemical reaction and set the ratio in which reactants and products combine. They are essential for converting between quantities in stoichiometric calculations.
How can moles be converted to the number of particles?
By multiplying the moles by the molar mass
By dividing the moles by the atomic mass
By multiplying the moles by Avogadro's number
By adding Avogadro's number to the moles
Moles are converted to the number of particles by multiplying by Avogadro's number (6.022 x 10^23). This conversion is fundamental in chemistry for linking the macroscopic and atomic scales.
Approximately how many particles are present in 2.5 moles of a substance?
1.51 x 10^24 particles
6.02 x 10^23 particles
2.5 x 10^24 particles
1.51 x 10^23 particles
Multiplying 2.5 moles by Avogadro's number (6.022 x 10^23) gives approximately 1.5055 x 10^24 particles, which rounds to about 1.51 x 10^24.
Given the reaction 2Al + 3Cl2 â†' 2AlCl3, if 50 g of Al reacts with excess Cl2, what is the mass of AlCl3 produced? (Molar masses: Al = 27 g/mol, AlCl3 ≈ 133.35 g/mol)
Approximately 300 grams
Approximately 150 grams
Approximately 246.7 grams
Approximately 350 grams
First, calculate the moles of Al by dividing 50 g by 27 g/mol, which gives roughly 1.85 moles. Since the reaction has a 1:1 mole ratio between Al and AlCl3, multiplying 1.85 moles by the molar mass of AlCl3 (133.35 g/mol) produces approximately 246.7 grams.
In the combustion reaction C3H8 + 5O2 â†' 3CO2 + 4H2O, burning 44 g of C3H8 produces how many grams of CO2? (Molar masses: C3H8 = 44 g/mol, CO2 = 44 g/mol)
132 grams
220 grams
176 grams
88 grams
Since 44 g of propane corresponds to 1 mole (given its molar mass is 44 g/mol), and the balanced equation indicates that 1 mole of C3H8 produces 3 moles of CO2, multiplying 3 moles by the molar mass of CO2 (44 g/mol) results in 132 grams of CO2.
For the reaction Fe2O3 + 3H2 â†' 2Fe + 3H2O, what is the theoretical yield of Fe when starting with 10.0 g of Fe2O3? (Molar masses: Fe2O3 ≈ 159.7 g/mol, Fe ≈ 55.85 g/mol)
Approximately 3.5 grams
Approximately 7.0 grams
Approximately 14.0 grams
Approximately 21.0 grams
Dividing 10.0 g of Fe2O3 by its molar mass (159.7 g/mol) gives about 0.0626 moles. Since the reaction produces 2 moles of Fe per mole of Fe2O3, the reaction yields approximately 0.1252 moles of Fe, which when multiplied by the molar mass of Fe (55.85 g/mol) results in roughly 7.0 grams.
In a two-step process, compound A converts to B with an 80% yield and then B converts to C with a 90% yield (both steps are 1:1 stoichiometrically). If 100 g of A is used, what is the mass of C produced?
100 grams
90 grams
80 grams
72 grams
The overall yield is the product of the individual yields: 80% (0.80) for the first step and 90% (0.90) for the second step, resulting in 0.80 x 0.90 = 0.72 or 72%. Thus, starting with 100 g of A yields 72 g of C.
When 10.0 g of potassium chlorate (KClO3) decomposes via 2KClO3 â†' 2KCl + 3O2, what is the volume of O2 produced at STP? (Molar mass of KClO3 ≈ 122.45 g/mol; STP volume = 22.4 L/mol)
Approximately 1.12 L
Approximately 4.12 L
Approximately 5.55 L
Approximately 2.74 L
First, calculate the moles of KClO3 by dividing 10.0 g by 122.45 g/mol, which is approximately 0.0817 moles. Using the stoichiometric ratio (2 moles KClO3 produce 3 moles O2), the moles of O2 produced is 0.0817 x (3/2) ≈ 0.12255 moles. Multiplying by the molar volume at STP (22.4 L/mol) gives roughly 2.74 L of O2.
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Study Outcomes

  1. Analyze balanced chemical equations to determine mole ratios.
  2. Apply the mole concept to convert between mass, moles, and number of particles.
  3. Calculate limiting reactants and theoretical yields in chemical reactions.
  4. Evaluate reaction efficiency using percent yield calculations.
  5. Solve multi-step stoichiometry problems with precise unit conversions.

Stoich Review Worksheet Cheat Sheet

  1. Understand the Mole Concept - Dive into the world of moles by memorizing that one mole equals 6.022 × 1023 particles. This lets you "count" atoms and molecules like a pro without listing each one. It's the chemist's secret sauce for converting microscopic particles into measurable quantities. Stoichiometry Overview at The Physics Classroom
  2. Master Balancing Chemical Equations - Get hands‑on practice adjusting coefficients so that each element has the same atom count on both sides of the arrow. This enforces the Law of Conservation of Mass and sharpens your problem‑solving instincts. Balanced equations are your roadmap for predicting product amounts. Balancing Equations at Purdue ChemEd
  3. Learn Mole Ratios - Use the coefficients from a balanced equation to find how many moles of one substance relate to another. This skill helps you scale reactions up or down without breaking a sweat. It's like having a built‑in recipe converter for chemical mixtures. Reaction Stoichiometry on OpenStax
  4. Convert Between Grams and Moles - Treat the molar mass as your conversion bridge between the mass of a substance and its mole count. Simply divide grams by grams-per-mole or multiply moles by grams-per-mole to switch back and forth. This keeps your calculations grounded in real‑world masses. Stoichiometry Resources at Marquis Tutoring
  5. Identify Limiting Reactants - Figure out which reactant runs out first and shuts down the reaction. The limiting reactant dictates the maximum amount of product you can make. Pinpointing it is key to error‑proof lab work and accurate yield predictions. Limiting Reactants Guide at The Physics Classroom
  6. Calculate Theoretical Yield - Predict the maximum product you could get under perfect conditions by using stoichiometry and the limiting reactant. This gives you a benchmark for what "should" happen before experimental hiccups. It's your ideal‑world target. Theoretical Yield Tips at Marquis Tutoring
  7. Determine Percent Yield - Compare your actual experimental yield to the theoretical yield and multiply by 100 to get a percent. This metric shows your reaction's efficiency and highlights any practical losses. Aim for higher percentages to boost your lab cred! Percent Yield Worksheet at Marquis Tutoring
  8. Understand Percent Composition - Calculate each element's mass percentage in a compound by dividing the total mass of the element by the compound's molar mass. This reveals the compound's makeup on a slice-by-slice basis. Use it to verify purity or determine formulas. Percent Composition Notes at Wilkes University
  9. Derive Empirical Formulas - Convert percent composition into moles, then simplify to the smallest whole‑number ratio. The empirical formula is like the compound's simplest recipe, showing only the essential ingredient proportions. It's foundational for identifying unknowns. Empirical Formula Exercises at Marquis Tutoring
  10. Apply Stoichiometry to Solutions - Use molarity (moles per liter) and solution volumes to calculate how much solute participates in a reaction. This approach handles titrations, dilutions, and all your liquid‑based experiments. It's the go‑to tool for solution chemistry. Solution Stoichiometry on OpenStax
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