Module 05: Stoichiometry Flashcards by carla klaasen

01.05 The Mole Concept

What is true about 1.0 mol Ca and 1.0 mol Mg? (3 points)

They are equal in mass.
They contain the same number of atoms.
They have the same atomic mass.
Their molar masses are equal.

2. They contain the same number of atoms.

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01.05 The Mole Concept

What is the molar mass of silver (Ag)? (3 points)

47.0 g/mol
60.02 g/mol
107.87 g/mol
196.97 g/mol

3. 107.87 g/mol

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01.05 The Mole Concept

How many moles of copper (Cu) are in 65.8 g Cu? (3 points)

1.04 mol Cu
10.9 mol Cu
41.7 mol Cu
63.5 mol Cu

1. 1.04 mol Cu

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01.05 The Mole Concept

How many moles are in a sample of 1.52 x 1024 atoms of mercury (Hg)? (3 points)

1.75 mol Hg
2.52 mol Hg
91.5 mol Hg
50.4 mol Hg

2. 2.52 mol Hg

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01.05 The Mole Concept

Avogadro’s number is used to determine the number of subatomic particles in an atom. (2 points)

True

False

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01.05 The Mole Concept

Which of the following processes will determine the number of moles in a sample? (3 points)

Dividing the mass of the sample by Avogadro’s number
Multiplying the mass of the sample by Avogadro’s number
Dividing the number of molecules in the sample by Avogadro’s number
Multiplying the number of molecules in the sample by Avogadro’s number

3. Dividing the number of molecules in the sample by Avogadro’s number

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01.05 The Mole Concept

How many moles of silver are equivalent to 2.408 × 1024 atoms? (3 points)

4
2.5
0.25
0.4

1. 4

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01.05 The Mole Concept

What is the number of atoms in a mole of any element? (3 points)

Avogadro’s number
Graham’s number
Its atomic number
Its mass number

1. Avogadro’s number

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01.05 The Mole Concept

What is the mass of 6.12 moles of arsenic (As)? (3 points)

12.2 g As
73.7 g As
276 g As
459 g As

4. 459 g As

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01.05 The Mole Concept

How many atoms of iron (Fe) are in a sample of 7.38 mol Fe? (3 points)

1.23 x 10²³ atoms Fe
5.58 x 10²³ atoms Fe
3.37 x 10²⁴ atoms Fe
4.44 x 10²⁴ atoms Fe

4. 4.44 x 10²⁴ atoms Fe

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01.05 The Mole Concept

Which of the following processes will determine the number of atoms in a sample? (3 points)

Dividing the mass of the sample by its molar mass
Multiplying the mass of the sample by its molar mass
Dividing the number of moles of the sample by Avogadro’s number
Multiplying the number of moles of the sample by Avogadro’s number

4. Multiplying the number of moles of the sample by Avogadro’s number

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01.05 The Mole Concept

How many moles of gold are equivalent to 1.204 × 1024 atoms? (3 points)

3. 2

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01.05 The Mole Concept

What is the empirical formula of C6H12O6? (4 points)

C2H4O2

CH2O

CH2O2

C2H4O

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01.05 The Mole Concept

What is a mole? What is the purpose?

Moles (mol): represent number & counting units

number of atoms in carbon 12 → avogadro’s number

Molar mass: total mass in grams on one mole in a substance

Amu on periodic table = 1 mole

Why use the mole?

many chemical properties depend on particle number (not mass)
Atoms too small count individually without special equipment
relationship between mass and number = determine numbe particles

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01.05 The Mole Concept

What is Avogadro’s Number?

Number particles in mole experimentally determined variety of ways

Use: mass spectrometer (count atoms precisely)

Avogadro’s number: 6.02214 × 10²³

1 mole = 6.022 × 10²³particles

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01.05 The Mole Concept

What is the relationship between moles and molar mass?

Mass of 1 Mole = Average Atomic Mass = Molar Mass

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01.05 The Mole Concept

How to convert from number of moles to number of atoms:

Given number of atoms x (6.022 x 10²³ atoms) / 1 mole = number of atoms

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01.05 The Mole Concept

How to convert from the number of atoms to moles?

Given number of atoms x (1 mole / 6.022 x 10²³ atoms) = number of moles

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01.05 The Mole Concept

How to convert from grams to atoms:

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How to convert from atoms to mass (g)?

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05.02 Molar Mass of Compounds

Define mass value on the periodic table:

mass (grams) of one mole (6.022 * 10²³) of an element

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05.02 Molar Mass of Compounds

How to calculate thee molar mass of a compound:

Subscrips: indicate amount of moles in compound

Step 01: Find molar mass on table

Step 02: multiply mass of each element by its subscript

Step 03: add up results

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05.02 Molar Mass of Compounds

What is the molar mass of Na₂CO₃? (3 points)

60.0 g/mol
106.0 g/mol
118.0 g/mol
141.0 g/mol

2. 106.0 g/mol

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05.02 Molar Mass of Compounds

How many moles of each element are in one mole of Be(OH)₂? (3 points)

1 mole of beryllium, 1 mole of oxygen, 2 moles of hydrogen
1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen
2 moles of beryllium, 2 moles of oxygen, 2 moles of hydrogen
2 moles of beryllium, 1 mole of oxygen, 1 moles of hydrogen

2. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen

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# 05.02 Molar Mass of Compounds How many moles are in 123.0 grams of KClO₄? (3 points) 1. 0.2354 mol KClO₄ 2. 0.6445 mol KClO₄ 3. 0.7124 mol KClO₄ 4. 0.8878 mol KClO₄

**4.** 0.8878 mol KClO₄

# 05.02 Molar Mass of Compounds What is the mass of 4.32 mol GaI₃? (3 points) 1. 455 g 2. 1,230 g 3. 1,720 g 4. 1,950 g

**4.** 1,950 g

# 05.02 Molar Mass of Compounds What is the molar mass of H₂SO₄? (Molar mass of H = 1.0079 g/mol; S = 32.065 g/mol; O = 15.999 g/mol) (3 points) 1. 98.08 g/mol 2. 88.22 g/mol 3. 86.79 g/mol 4. 79.02 g/mol

**1.** 98.08 g/mol

# 05.02 Molar Mass of Compounds Calculate the mass of 3.4 moles of nitric acid (HNO₃). Explain the process or show your work by including all values used to determine the answer. (5 points)

Firstly, determine the molar mass of nitric acid. There is 1 atom of Hydrogen, 1 atom of Nitrogen, and three atoms of Oxygen. Determine their individual masses and add the multiple of each other to find the total molar mass of the compound: 1H+1N+3O=1(1.0079)+1(14.0067)+3(15.9994)=63.0128 Therefore, the molar mass is 63.0128 g/mol HNO3. Next, in order to determine the mass of 3.4 of HNO3, an appropriate conversion factor needs to be found. Since the conversion is from moles to mass, the following formula is needed: Given moles of a sample \* (molar mass of compound) ÷ (1 mole) = mass of a sample. Hence, the conversion is 3.4 mol \* (63.0128 g/mol)/1 mol = x. 3.4⋅63.01281=214.24g. Therefore, the mass of 3.4 moles of nitric acid is 214.24 grams.

# 05.03 The Empirical Formula What is Percent Composition?

* Ratios * represented empirical formula & molar mass * * * **Step 01: Find the number of atoms in the empirical formula** **Step 02: Find the molar mass** 1. find on periodic table 2. multiply value in empirical formula **Step 03: Total individual molar masses to find total** **Step 04: Calculate Percentage Composition by divide the molar mass of each element by molar mass of total molar mass**

# 05.03 The Empirical Formula Determine Empirical Formula:

**_Step 01_: Check composition of compound (should be in grams)** * *if the sample is in percentage convert it assuming it is a 100-gram sample* **_Step 02_: Convert amount element from grams to moles** * *conversion factor of (1 mole) ÷ (molar mass of substance)* **_Step 03:_ Divide each mole value by lowest mole value of all elements = whole-number mole ratio** * mole values are mole ratio * empirical formula requires lowest whole-number ratio **_Step 04:_ If the values are not a whole number, multiply all mole values by a number make them a whole number** **_Step 05:_ Use whole-numbers as subscripts when writing the empirical formula**

# 05.03 The Empirical Formula How to convert from the Empirical formula to the Molecular formula?

**_Step 01_: Calculate molar mass of empirical formula in same way you would calculate molar mass of a compound** **_Step 02_: Divide (entire compound) molecular mass by mass empirical formula → ratio between molecular formula and empirical formula (whole number)** **_Step 03_: Multiply subscripts in empirical formula by ratio to get molecular formula**

# 05.03 The Empirical Formula What is the empirical formula of C₆H₁₂O₆? (4 points) 1. C₂H₄O₂ 2. CH₂O 3. CH₂O₂ 4. C₂H₄O

**2.** CH₂O

# 05.03 The Empirical Formula Which pair shares the same empirical formula? (4 points) 1. C2H4 and C6H6 2. C6H6 and C3H3 3. CH2 and C6H6 4. CH and C2H4

**2.** C₆H₆ and C₃H₃

# 05.03 The Empirical Formula What is the percent composition of NaHCO₃? (4 points) 1. 23.20% Na, 4.26% H, 18.41% C, and 54.13% O 2. 24.21% Na, 4.35% H, 12.26% C, and 59.18% O 3. 27.36% Na, 1.20% H, 14.30% C, and 57.14% O 4. 30.44% Na, 2.12% H, 10.25% C, and 60.19% O

**3.** 27.36% Na, 1.20% H, 14.30% C, and 57.14% O

# 05.03 The Empirical Formula A compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula? (4 points) 1. CH₂O₂ 2. CH₂O 3. C₂H₄O 4. C₂H₄O₂

**2.** CH₂O

# 05.03 The Empirical Formula The empirical formula of a chemical substance is CH. The molar mass of a molecule of the substance is 78.11 g/mol. What is the molecular formula of the chemical substance? (4 points) 1. C2H2 2. C2H4 3. C3H4 4. C6H6

**4.** C6H6

# 05.04 Stoichiometry Define stoichiometry:

dimensional analysis (math conversions) starts with one substance and ends with different substance

# 05.04 Stoichiometry What is a mole ratio?

a conversion factor used in stoichiometry that describes the ratio between compounds or elements in a chemical reaction

# 05.04 Stoichiometry How do you determine the mass value of a substance?

Step 01**: Write a balanced Equation** Step 02**: Begin stoichiometry using molar mass of starting substance** Step 03**: Create mole ratio from coefficients of two compounds of interest** Step 04: **Include mole ratio in stoichiometry conversion** Step 05: **Include conversion factor for the molar mass of ending substance** Step 06: **Solve through dimensional analysis**

# 05.04 Stoichiometry What is a valid mole ratio from the balanced equation 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O? (4 points)

6 mole H20 / 9 mole O2

# 05.04 Stoichiometry How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)? 2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃₎₂ (4 points) 1. 2.2 mol CaCl2 2. 3.1 mol CaCl2 3. 6.2 mol CaCl2 4. 12.4 mol CaCl2

**2.** 3.1 mol CaCl2

# 05.04 Stoichiometry What mass of Fe can be produced by the reaction of 75.0 g CO with excess Fe₂O₃ according to the equation: Fe2O₃(s) + CO(g) → Fe(s) + CO₂(g)? (4 points) 1. 49.8 g Fe 2. 99.7 g Fe 3. 224 g Fe 4. 299 g Fe

**2.** 99.7 g Fe

# 05.04 Stoichiometry How many moles of sulfuric acid (H2SO4) are needed to react completely with 6.8 moles of lithium hydroxide (LiOH)? 2LiOH + H₂SO₄ → Li₂SO₄+ 2H₂O (4 points) 1. 3.4 mol H2SO4 2. 6.8 mol H2SO4 3. 10.2 mol H2SO4 4. 13.6 mol H2SO4

**1.** 3.4 mol H2SO4

# 05.04 Stoichiometry What mass of Fe can be produced by the reaction of 75.0 g CO with excess Fe2O3 according to the equation: Fe2O3(s) + CO(g) → Fe(s) + CO2(g)? (4 points) 1. 49.8 g Fe 2. 99.7 g Fe 3. 224 g Fe 4. 299 g Fe

**2.** 99.7 g Fe

# 05.04 Stoichiometry In an experiment, potassium chlorate decomposed according to the following chemical equation. **KClO₃ → KCl + O₂** (Molar mass of KClO3 = 122.5 g/mol; KCl = 74.55 g/mol; O2 = 31.998 g/mol) If the mass of potassium chlorate was 240 grams, which of the following calculations can be used to determine the mass of oxygen gas formed? (4 points) 1. (240 × 2 × 31.998) ÷ (122.5 × 3) grams 2. (240 × 3 × 31.998) ÷ (122.5 × 2) grams 3. (240 × 2 × 122.5) ÷ (31.998 × 3) grams 4. (240 × 3 × 122.5) ÷ (31.998 × 2) grams

**2.** (240 × 3 × 31.998) ÷ (122.5 × 2) grams

# 05.05 Limiting Reactant Define limiting and excess reactant:

_Limiting Reactant_: Runs out first _Excess Reactant_: not used up completely

# 05.05 Limiting Reactant Define Theoretical Yield:

maximum amount product expect from reaction (one run out before the other)

# 05.05 Limiting Reactant What are the steps to determine the limiting reactants and moles?

**Step 01**: Determine limiting reactant by identifying the given about of reactants and what product is being solved for **Step 02:** Write balanced equations and use it to create a mole ratio for the given reactant and product **Step 03:** Set up and solve a stoichiometry calculation for each reactant using the mole ratios **Step 04:** The Lower amous is the theoretical yield

# 05.05 Limiting Reactant How to determine the mass of a theoretical yield?

Step 01: Determine if this is a limiting reactant problem by recognizing there are given starting amounts for both reactants. Balance the chemical reaction. Step 02: Set up the stoichiometry calculation for each reactant using the periodic table to determine individual moral mass Step 03: Use balanced equation to create mole ratio for the given reactants and products Step 04: Include mole ratio for each stoichiometry calculation Step 05: Include conversion factor for molar mass of ending substance for each calculation (solve through dimensional analysis) Step 06: Lower amount is theoretical yield

# 05.05 Limiting Reactant Consider the reaction 2CuCl₂ + 4KI → 2CuI + 4KCl + I2. If 4 moles of CuCl2 react with 4 moles of KI, what is the limiting reactant? (3 points) 1. CuCl2 2. KI 3. CuI 4. I2

2. Kl

# 05.05 Limiting Reactant If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed? CuCl₂ + NaNO₃ → Cu(NO₃)₂ + NaCl (3 points) 1. 12.4 g NaCl 2. 15.7 g NaCl 3. 16.2 g NaCl 4. 28.4 g NaCl

**2.** 15.7 g NaCl

# 05.05 Limiting Reactant Read the chemical equation. Fe₂O₃ + CO → Fe + CO₂ If 3 moles of Fe₂O₃react with 1.5 moles of CO, how many moles of each product are formed? (3 points) 1. 1 mole of Fe and 1.5 moles of CO2 2. 0.5 mole of Fe and 1 mole of CO2 3. 6 moles of Fe and 9 moles of CO2 4. 3 moles of Fe and 2 moles of CO2

**1,** 1 mole of Fe and 1.5 moles of CO2

# 05.06 Percent Yield Define percentage yield:

**Percent Yield: ^{Actual Yield} / _{Theoretical Yield} \* 100** Percent Yield: _ration acual yield to theoretical yield multiplied by 100_ **Theoretical Yield**: maximum amount of product able to be produced from a given amount of reactant(s) (calculated) **Actual Yield**: amount of product that is actually formed (measured) _reasons_: 1. conditions not allowed reaction run to completion 2. purification of product results in loss of some of the product *The actual yield cannot exceed the theoretical yield. In practice, sometimes it appears that way, but that's because an error occurred in the lab. This will be important to remember as you complete percent yield calculations.*

# 05.09 Honors Stoichiometry Exam 05.09 Honors Stoichiometry Exam

# 05.09 Honors Stoichiometry Exam How many atoms of Mg are present in 97.22 grams of Mg? (4 points) 1. 6.022 × 10²³ 2. 2.408 × 10²⁴ 3. 4.818 × 10²⁴ 4. 5.855 × 10²⁵

**2.** 2.408 × 10²⁴

# 05.09 Honors Stoichiometry Exam One mole of zinc has a mass of 65.4 grams. Approximately how many atoms of zinc are present in one mole of zinc? (4 points) 1. 32 × 10²³ atoms 2. 6 × 10²³ atoms 3. 66 atoms 4. 65 atoms

**2.** 6 × 10²³ atoms

# 05.09 Honors Stoichiometry Exam Which of the following best defines the molar mass of a substance? (4 points) 1. Mass, in grams, of each particle of the substance 2. Total mass, in grams, of 100 particles of the substance 3. Mass, in grams, of the nucleus of the substance's atoms 4. Total mass, in grams, of all the particles in one mole of the substance

**4.** Total mass, in grams, of all the particles in one mole of the substance

# 05.09 Honors Stoichiometry Exam Which formula can be used to calculate the molar mass of ammonia (NH3)? (4 points) 1. molar mass of N + molar mass of H 2. 3 × molar mass of N + molar mass of H 3. molar mass of N + 3 × molar mass of H 4. 3 × molar mass of N + 3 × molar mass of H

**3.** molar mass of N + 3 × molar mass of H

# 05.09 Honors Stoichiometry Exam How many moles of MgCO3 are present in 252.939 grams of MgCO3? (4 points) 1. 2 2. 3 3. 5 4. 6

**2.** 3

# 05.09 Honors Stoichiometry Exam Which pair of compounds has the same empirical formula? (4 points) 1. C2H6 and CH 2. CH3 and C2H6 3. C3H8 and C3H 4. C2H2 and C2H

**2.** CH3 and C2H6

# 05.09 Honors Stoichiometry Exam What is the percentage composition of each element in hydrogen peroxide, H2O2? (4 points) 1. 7.01% H and 92.99% O 2. 7.22% H and 92.78% O 3. 6.32% H and 93.68% O 4. 5.88% H and 94.12% O

**4.** 5.88% H and 94.12% O

# 05.09 Honors Stoichiometry Exam Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound? For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (8 points)

**Empirical Formula:** Convert to moles: 47. 37 grams of carbon = 3.944 moles of carbon 10. 58 grams of hydrogen = 10.497 moles of hydrogen 42. 04 grams of oxygen = 2.628 moles of oxygen Divide each element by the lowest value: 3. 944 mol C ÷ 2.628 = 1.500 mol C 10. 497 mol H ÷ 2.628 = 3.994 = 4 mol H 2. 628 mol O ÷ 2.628 = 1 mole O 1. 5 \* 2 = 3 mol C 4 \* 2 = 8 mol H 1 \* 2 = 2 mol O Therefore, the empirical formula is **C3H8O2**. **Molecular Formula:** Molar mass of molecule: _228.276 g/mol_ Molar mass empirical formula: 3(12.0107) + 8(1.0079) + 2(15.9994) = _76.0941 g/mol_ =228.27676.0941 =2.999≈3 =3⋅C3H8O2 =C9H24O6

# 05.09 Honors Stoichiometry Exam Which of the following is necessary to do a complete stoichiometric calculation? (4 points) 1. Write the mole ratio. 2. Add the nuclear masses of the products. 3. Divide the nuclear masses of the reactants. 4. Write the number of atoms in one mole of the product.

**1.** Write the mole ratio

# 05.09 Honors Stoichiometry Exam The following reaction shows sodium hydroxide reacting with sulfuric acid. **4NaOH + 2H₂SO₄ → 2Na₂SO₄ + 4H₂O** How many grams of Na2SO4 are produced from 10.0 grams of NaOH? (Molar mass of Na = 22.989 g/mol, O = 15.999 g/mol, H = 1.008 g/mol, S = 32.065 g/mol) (4 points) 1. 17.8 grams 2. 19.2 grams 3. 35.5 grams 4. 38.5 grams

**1.** 1. 17.8 grams

# 05.09 Honors Stoichiometry Exam Read the given chemical reaction. **C₂H₆ + O₂ → CO₂ + H₂O** How many moles of CO2 are produced during the complete combustion of 3.6 moles of C2H6? (4 points) 1. 1.8 moles 2. 4.4 moles 3. 7.2 moles 4. 9.2 moles

**3.** 7.2 moles

# 05.09 Honors Stoichiometry Exam Which of the following statements best describes a limiting reactant? (4 points) 1. The reactant that is used up the least 2. The reactant that is used up the most 3. The reactant that is used up before the others 4. The reactant that is leftover after the reaction

**3.** The reactant that is used up before the others

# 05.09 Honors Stoichiometry Exam How many moles of water are produced when 3.0 moles of hydrogen gas react with 1.8 moles of oxygen gas? Balanced equation: **2H₂ + O₂ → 2H₂O** (4 points) 1. 3.0 moles of water 2. 3.6 moles of water 3. 5.3 moles of water 4. 7.0 moles of water

**1.** 3.0 moles of water

# 05.09 Honors Stoichiometry Exam A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solution of 0.40 grams of copper (II) chloride. A single replacement reaction takes place. What are the likely observations when the reaction stops? Unbalanced equation: **CuCl₂ + Al → AlCl₃ + Cu** (4 points) 1. About 0.90 grams of copper is formed, and some aluminum is left in the reaction mixture 2. About 0.20 grams of copper is formed, and some aluminum is left in the reaction mixture. 3. About 0.90 grams of copper is formed, and some copper chloride is left in the reaction mixture. 4. About 0.20 grams of copper is formed, and some copper chloride is left in the reaction mixture.

**2.** About 0.20 grams of copper is formed, and some aluminum is left in the reaction mixture

# 05.09 Honors Stoichiometry Exam Which of the following statements best defines the theoretical yield of a reaction? (4 points) 1. The ratio of measured yield over actual yield 2. The amount of product measured after a reaction 3. The ratio of measured yield over stoichiometric yield 4. The maximum amount of product that can be obtained

**4.** The maximum amount of product that can be obtained

# 05.09 Honors Stoichiometry Exam The actual yield of a product in a reaction was measured as 4.20 g. If the theoretical yield of the product for the reaction is 4.88 g, what is the percentage yield of the product? (4 points) 82. 6% 84. 2% 86. 1% 88. 0%

3. 86.1 %

# 05.09 Honors Stoichiometry Exam A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown. **Na + Cl₂ → NaCl** If the percentage yield of the reaction is 86%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors. (8 points)

**Theoretical Yield:** Since the chlorine gas in excess, sodium is the limiting reactant in the reaction. The balanced equation for the above reaction is 2Na+Cl2→2NaCl. The theoretical yield will be as follows: =57.51⋅145.9794⋅22⋅116.88541 =146.172 grams theoretical yield **Actual Yield:** Actual Yield = (Percent Yield \* Theoretical Yield) ÷ 100 =12570.792100 =125.708100 =125.708 grams actual yield

# 05.09 Honors Stoichiometry Exam The following reaction shows the products when sulfuric acid and aluminum hydroxide react. **2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O** The table shows the calculated amounts of reactants and products when the reaction was conducted in a laboratory. What is the approximate amount of the leftover reactant? (4 points) 1. 8.2 g of sulfuric acid 2. 9.8 g of sulfuric acid 3. 11.43 g of aluminum hydroxide 4. 13.76 g of aluminum hydroxide

**3.** 11.43 g of aluminum hydroxide

# 05.09 Honors Stoichiometry Exam In the following reaction, oxygen is the excess reactant. **SiCl₄ + O₂ → SiO₂ + Cl₂** The table shows an experimental record for the above reaction. ## Footnote *Calculate the percentage yield for SiO2 for Trial 1. Also, determine the leftover reactant for the trial. Show your work.*

In the following reaction, oxygen is the excess reactant. SiCl4 + O2 → SiO2 + Cl2 The table shows an experimental record for the above reaction. ## Footnote Based on the percentage yield in Trial 2, explain what ratio of reactants is more efficient for the given reaction.