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Chapter 11: Problem 37

What relationships can be determined from a balanced chemical equation?

Short Answer

Expert verified
From a balanced chemical equation, we can determine stoichiometric coefficients, mass relationships, and mole-to-mole relationships between reactants and products. For example, in the combustion of methane (\(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\)), the stoichiometric coefficients show that 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O. The mass relationships indicate that 16.05 g of CH4 reacts with 64.00 g of O2 to produce 44.01 g of CO2 and 36.04 g of H2O.

Step by step solution

01

Pick a simple balanced chemical equation

Choose a simple balanced chemical equation that illustrates all the relationships. For this exercise, we will use the combustion of methane (CH4) as an example. The balanced chemical equation for the combustion of methane is: \[CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\]
02

Identify stoichiometric coefficients

Stoichiometric coefficients are the numerical values that appear in front of the chemical formulas in a balanced chemical equation. They determine the ratio of reactants to products in the equation. In our example: 1. The coefficient of CH4 is 1 2. The coefficient of O2 is 2 3. The coefficient of CO2 is 1 4. The coefficient of H2O is 2 This means that for every 1 mole of methane reacted, it requires 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
03

Show mass relationships

The mass relationships between the reactants and products in a balanced chemical equation can also be determined. To do this, we must first calculate the molar mass of each species in the equation. Using a periodic table: 1. Molar mass of CH4 = 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol 2. Molar mass of O2 = 2 * 16.00 g/mol (O) = 32.00 g/mol 3. Molar mass of CO2 = 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol 4. Molar mass of H2O = 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol The mass relationships can now be determined from the balanced equation. For every 16.05 g of CH4 reacted, 64.00 g of O2 is required to produce 44.01 g of CO2 and 36.04 g of H2O.
04

Find mole relationships

The balanced chemical equation also shows the mole-to-mole relationships between the reactants and products. Using the stoichiometry and the molar mass relationships, we can say that: 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O. These relationships can be used to predict the quantities of reactants and products involved in a chemical reaction based on the balanced chemical equation. In conclusion, a balanced chemical equation can help identify stoichiometric coefficients, mass relationships, and mole-to-mole relationships between reactants and products, which are essential in determining the quantities and ratios of substances involved in a chemical reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometric Coefficients
In every balanced chemical equation, you'll see numbers written before the chemical formulas. These are called stoichiometric coefficients. They aren't just arbitrary numbers; they play a crucial role in determining the exact ratio of substances involved in the reaction. In the combustion of methane (\(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\)):
  • The coefficient of methane (\(CH_4\)) is 1.
  • Oxygen (\(O_2\)) has a coefficient of 2.
  • Carbon dioxide (\(CO_2\)) has a coefficient of 1.
  • And water (\(H_2O\)) has a coefficient of 2.
What does this mean? For every 1 mole of methane burned, you need 2 moles of oxygen. This results in 1 mole of carbon dioxide and 2 moles of water produced. By understanding these coefficients, you can predict how much reactant you need and how much product you'll get. This is vital for experiments and industrial applications.
Mass Relationships
Mass relationships in a balanced chemical equation help connect the world of moles to the tangible world of grams and kilograms. They ensure that when dealing with real substances, everything remains consistent and accountable. To uncover mass relationships, we first find the molar mass of each reactant and product:
  • Methane (\(CH_4\)) has a molar mass of 16.05 g/mol.
  • Oxygen (\(O_2\)) has a molar mass of 32.00 g/mol.
  • Carbon dioxide (\(CO_2\)) has a molar mass of 44.01 g/mol.
  • Water (\(H_2O\)) has a molar mass of 18.02 g/mol.
From the balanced equation \(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\), we find that for burning 16.05 grams of methane, 64.00 grams of oxygen is needed. This process produces 44.01 grams of carbon dioxide and 36.04 grams of water. Understanding these mass relationships is essential for practical applications like measuring chemicals before conducting reactions.
Mole-to-Mole Relationships
Mole-to-mole relationships are at the heart of stoichiometry, which uses the language of moles to translate the chemical equation into something we can calculate and use. For the methane combustion reaction:
  • 1 mole of \(CH_4\) reacts with 2 moles of \(O_2\).
  • The reaction yields 1 mole of \(CO_2\) and 2 moles of \(H_2O\).
These mole-to-mole ratios directly relate how much of each substance is involved in the reaction. If you start with 2 moles of methane, you'll need 4 moles of oxygen, yielding 2 moles of carbon dioxide and 4 moles of water. By understanding these ratios, students can solve problems involving unknown quantities, predict yields, and understand the dynamics of reactions at the molecular level.
Stoichiometry thus provides the tools to confidently navigate and solve chemical equations, ensuring accurate and successful reactions every time.

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Most popular questions from this chapter

One of the reactions used to inflate automobile air bags involves sodium azide \(\left(\mathrm{NaN}_{3}\right) : 2 \mathrm{NaN}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_{2}(\mathrm{g})\) mass of \(\mathrm{N}_{2}\) produced from the decomposition of NaN \(_{3}\) shown at right.

Why are coefficients used in mole ratios instead of subscripts?

Describe how a mole ratio is correctly expressed when it is used to solve a stoichiometric problem.

Car Battery Car batteries use lead, lead(IV) oxide, and a sulfuric acid solution to produce an electric current. The products of the reaction are lead(II) sulfate in solution and water. a. Write the balanced equation for the reaction. b. Determine the mass of lead(II) sulfate produced when 25.0 g of lead reacts with an excess of lead(IV) oxide and sulfuric acid.

Challenge Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride \(\left(\mathrm{TiCl}_{4}\right)\) is extracted from titanium oxide \(\left(\mathrm{TiO}_{2}\right)\) using chlorine and coke (carbon). $$\mathrm{TiO}_{2}(\mathrm{s})+\mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{TiCl}_{4}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$$ a. What mass of \(\mathrm{Cl}_{2}\) gas is needed to react with 1.25 \(\mathrm{mol}\) of \(\mathrm{TiO}_{2} ?\) b. What mass of \(C\) is needed to react with 1.25 mol of TiO \(_{2} ?\) c. What is the mass of all of the products formed by reaction with 1.25 \(\mathrm{mol}\) of \(\mathrm{TiO}_{2} ?\)
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