Domanda del libro di testo
Balance the following equation, and tell how many moles of nickel will react with 9.81 mol of hydrochloric acid.
Ni(s) + HCl(aq) → NiCl2(aq) + H2(g)
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Ch.6 Chemical Reactions: Mole and Mass Relationships
McMurry8th EditionFundamentals of General, Organic, and Biological ChemistryISBN: 9780134015187
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TOC.tableOfContents
- Ch.1 Matter and Measurements27
- Ch.2 Atoms and the Periodic Table20
- Ch.3 Ionic Compounds8
- Ch.4 Molecular Compounds23
- Ch.5 Classification & Balancing of Chemical Reactions12
- Ch.6 Chemical Reactions: Mole and Mass Relationships28
- Ch.7 Chemical Reactions: Energy, Rate and Equilibrium64
- Ch.8 Gases, Liquids and Solids24
- Ch.9 Solutions52
- Ch.10 Acids and Bases25
- Ch.11 Nuclear Chemistry22
- Ch.12 Introduction to Organic Chemistry: Alkanes57
- Ch.13 Alkenes, Alkynes, and Aromatic Compounds47
- Ch.14 Some Compounds with Oxygen, Sulfur, or a Halogen50
- Ch.15 Aldehydes and Ketones45
- Ch.16 Amines42
- Ch.17 Carboxylic Acids and Their Derivatives57
- Ch.18 Amino Acids and Proteins82
- Ch.19 Enzymes and Vitamins63
- Ch.20 Carbohydrates44
- Ch.21 The Generation of Biochemical Energy65
- Ch.22 Carbohydrate Metabolism45
- Ch.23 Lipids42
- Ch.24 Lipid Metabolism33
- Ch.25 Protein and Amino Acid Metabolism24
- Ch.26 Nucleic Acids and Protein Synthesis45
McMurry8th EditionFundamentals of General, Organic, and Biological ChemistryISBN: 9780134015187Non è quello che usi tu?Cambia libro di testo
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McMurry 8th EditionCh.6 Chemical Reactions: Mole and Mass RelationshipsProblema 13
McMurry 8th EditionCh.6 Chemical Reactions: Mole and Mass RelationshipsProblema 13Capitolo 6, Problema 13
The reaction of ethylene oxide with water to give ethylene glycol (automobile antifreeze) occurs in 96.0% actual yield. How many grams of ethylene glycol are formed by reaction of 35.0 g of ethylene oxide? (For ethylene oxide, MW = 44.0 amu; for ethylene glycol, MW = 62.0 amu.)
Guida verificata passo dopo passo1
Step 1: Write the balanced chemical equation for the reaction. Ethylene oxide (C₂H₄O) reacts with water (H₂O) to form ethylene glycol (C₂H₆O₂). The balanced equation is: C₂H₄O + H₂O → C₂H₆O₂.
Step 2: Calculate the number of moles of ethylene oxide (C₂H₄O) using its molecular weight. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \). Substitute the given mass of ethylene oxide (35.0 g) and its molecular weight (44.0 g/mol).
Step 3: Use the stoichiometry of the reaction to determine the moles of ethylene glycol (C₂H₆O₂) produced. From the balanced equation, the molar ratio of ethylene oxide to ethylene glycol is 1:1. Therefore, the moles of ethylene glycol formed will be equal to the moles of ethylene oxide reacted.
Step 4: Convert the moles of ethylene glycol to grams using its molecular weight. Use the formula: \( \text{mass} = \text{moles} \times \text{molecular weight} \). Substitute the moles of ethylene glycol and its molecular weight (62.0 g/mol).
Step 5: Account for the actual yield of the reaction. Multiply the theoretical mass of ethylene glycol (calculated in Step 4) by the actual yield (96.0%) expressed as a decimal (0.960). This will give the actual mass of ethylene glycol produced.
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Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows us to determine the amount of product formed from a given amount of reactant by using molar ratios derived from the coefficients in the balanced equation.
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Percorso guidato
01:16
Stoichiometry
Molar Mass
Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is essential for converting between grams and moles, enabling the calculation of how many moles of a substance are present in a given mass, which is crucial for stoichiometric calculations.
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Percorso guidato
01:10Calculating Molar Mass
Percent Yield
Percent yield is a measure of the efficiency of a chemical reaction, calculated as the ratio of the actual yield to the theoretical yield, multiplied by 100. It indicates how much of the expected product was actually obtained, which is important for assessing the success of a reaction and for practical applications in chemistry.
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Percorso guidato
03:09Percent Yield
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