Question

For each of the following balanced chemical cquations, calculate how many moles of product(s) would be produced if 0.500 mole of the first reactant were to react completely. a. \(\mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{BaCl}_{2}(a q)+2 \mathrm{AgNO}_{3}(a q) \rightarrow 2 \mathrm{AgCl}(s)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{CO}_{2}(g)\) d. \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{Fe}(s) \rightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2}(g)\)

Short answer

a. \(0.500 \, \text{moles} \, \mathrm{CH}_{4}\) and \(1.00 \, \text{moles} \, \mathrm{H}_{2} \mathrm{O}\) b. \(1.00 \, \text{moles} \, \mathrm{AgCl}\) and \(0.500 \, \text{moles} \, \mathrm{Ba(NO}_{3}\mathrm{)_2}\) c. \(2.00 \, \text{moles} \, \mathrm{H}_{2} \mathrm{O}\) and \(1.50 \, \text{moles} \, \mathrm{CO}_{2}\) d. \(0.167 \, \text{moles} \, \mathrm{Fe}_{2}(SO}_{4}\mathrm{)_3\) and \(0.500 \, \text{moles} \, \mathrm{H}_{2}\)

Step-by-step solution

Identify the given reactant and the products

In the given equation: \(\mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g)+2\mathrm{H}_{2} \mathrm{O}(l)\) The given reactant is \(\mathrm{CO}_{2}\), and the products are \(\mathrm{CH}_{4}\) and \(\mathrm{H}_{2}\mathrm{O}\).

Apply stoichiometry

For every mole of \(\mathrm{CO}_{2}\), we have one mole of \(\mathrm{CH}_{4}\) produced and two moles of \(\mathrm{H}_{2}\mathrm{O}\). From the given amount (0.500 moles) of \(\mathrm{CO}_{2}\), we can calculate the moles of both products produced: For \(\mathrm{CH}_{4}\): \(0.500 \, \text{moles} \, \mathrm{CO}_{2} \times \frac{1 \, \text{mole} \, \mathrm{CH}_{4}}{1 \, \text{mole} \, \mathrm{CO}_{2}} = \boxed{0.500 \, \text{moles} \, \mathrm{CH}_{4}}\) For \(\mathrm{H}_{2}\mathrm{O}\): \(0.500 \, \text{moles} \, \mathrm{CO}_{2} \times \frac{2 \, \text{moles} \, \mathrm{H}_{2}\mathrm{O}}{1 \, \text{mole} \, \mathrm{CO}_{2}} = \boxed{1.00 \, \text{moles} \, \mathrm{H}_{2} \mathrm{O}}\) #Problem b#

Identify the given reactant and the products

In the given equation: \(\mathrm{BaCl}_{2}(a q)+2 \mathrm{AgNO}_{3}(a q) \rightarrow 2 \mathrm{AgCl}(s)+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) The given reactant is \(\mathrm{BaCl}_{2}\), and the products are \(\mathrm{AgCl}\) and \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\).

Apply stoichiometry

For every mole of \(\mathrm{BaCl}_{2}\), we have two moles of \(\mathrm{AgCl}\) produced and one mole of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\). From the given amount (0.500 moles) of \(\mathrm{BaCl}_{2}\), we can calculate the moles of both products produced: For \(\mathrm{AgCl}\): \(0.500 \, \text{moles} \, \mathrm{BaCl}_{2} \times \frac{2 \, \text{moles} \, \mathrm{AgCl}}{1 \, \text{mole} \, \mathrm{BaCl}_{2}} = \boxed{1.00 \, \text{moles} \, \mathrm{AgCl}}\) For \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\): \(0.500 \, \text{moles} \, \mathrm{BaCl}_{2} \times \frac{1 \, \text{mole} \, \mathrm{Ba(NO}_{3}\mathrm{)_2}}{1 \, \text{mole} \, \mathrm{BaCl}_{2}} = \boxed{0.500 \, \text{moles} \, \mathrm{Ba(NO}_{3}\mathrm{)_2}}\) #Problem c#

Identify the given reactant and the products

In the given equation: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{CO}_{2}(g)\) The given reactant is \(\mathrm{C}_{3} \mathrm{H}_{8}\), and the products are \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{CO}_{2}\).

Apply stoichiometry

For every mole of \(\mathrm{C}_{3} \mathrm{H}_{8}\), we have four moles of \(\mathrm{H}_{2}\mathrm{O}\) produced and three moles of \(\mathrm{CO}_{2}\). From the given amount (0.500 moles) of \(\mathrm{C}_{3} \mathrm{H}_{8}\), we can calculate the moles of both products produced: For \(\mathrm{H}_{2}\mathrm{O}\): \(0.500 \, \text{moles} \, \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{4 \, \text{moles} \, \mathrm{H}_{2}\mathrm{O}}{1 \, \text{mole} \, \mathrm{C}_{3} \mathrm{H}_{8}} = \boxed{2.00 \, \text{moles} \, \mathrm{H}_{2} \mathrm{O}}\) For \(\mathrm{CO}_{2}\): \(0.500 \, \text{moles} \, \mathrm{C}_{3} \mathrm{H}_{8} \times \frac{3 \, \text{moles} \, \mathrm{CO}_{2}}{1 \, \text{mole} \, \mathrm{C}_{3} \mathrm{H}_{8}} = \boxed{1.50 \, \text{moles} \, \mathrm{CO}_{2}}\) #Problem d#

Identify the given reactant and the products

In the given equation: \(3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{Fe}(s) \rightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2}(g)\) The given reactant is \(\mathrm{H}_{2} \mathrm{SO}_{4}\), and the products are \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) and \(\mathrm{H}_{2}\).

Apply stoichiometry

For every three moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), we have one mole of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) produced and three moles of \(\mathrm{H}_{2}\). From the given amount (0.500 moles) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), we can calculate the moles of both products produced: For \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\): \(0.500 \, \text{moles} \, \mathrm{H}_{2} \mathrm{SO}_{4} \times \frac{1 \, \text{mole} \, \mathrm{Fe}_{2}(SO}_{4}\mathrm{)_3}{3 \, \text{moles} \, \mathrm{H}_{2} \mathrm{SO}_{4}} = \boxed{0.167 \, \text{moles} \, \mathrm{Fe}_{2}(SO}_{4}\mathrm{)_3}\) For \(\mathrm{H}_{2}\): \(0.500 \, \text{moles} \, \mathrm{H}_{2} \mathrm{SO}_{4} \times \frac{3 \, \text{moles} \, \mathrm{H}_{2}}{3 \, \text{moles} \, \mathrm{H}_{2} \mathrm{SO}_{4}} = \boxed{0.500 \, \text{moles} \, \mathrm{H}_{2}}\)

Key concepts

Chemical Equations

Understanding chemical equations is fundamental in the study of chemistry. A chemical equation is a symbolic representation of a chemical reaction, showcasing the reactants that interact to form products. Each equation is carefully balanced, meaning the number of atoms for each element is conserved from reactants to products. This balance is crucial and follows the Law of Conservation of Mass.

For example, consider the simple equation \(\mathrm{CO}_2(g) + 4 \mathrm{H}_2(g) \rightarrow \mathrm{CH}_4(g) + 2 \mathrm{H}_2O(l)\). Here, the reactant carbon dioxide (CO2) reacts with hydrogen gas (H2) to produce methane (CH4) and water (H2O). The balancing numbers, also known as stoichiometric coefficients, in front of each molecule, indicate the relative amounts of reactants that are consumed and the amounts of products that are formed during the reaction.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure in the laboratory. One mole is defined as the amount of a substance that contains as many entities (atoms, molecules, ions, or other particles) as there are atoms in 12 grams of pure carbon-12. This number, Avogadro's number, is approximately \(6.022 \times 10^{23}\) entities per mole.

Understanding the mole allows chemists to count particles by weighing them. Knowing that 1 mole of a substance contains a definite number of particles, we can use this information to convert between mass, number of moles, and number of particles—enabling us to perform stoichiometric calculations and understand the quantitative aspects of reactions.

Reactants and Products

In a chemical reaction, reactants are the starting materials, the substances that undergo chemical changes. Meanwhile, the products are the substances that are produced as a result of the reaction. The ability to identify reactants and products in the chemical equation is vital since this sets the stage for any calculation or analysis of the reaction.

Different types of substances such as gases (g), liquids (l), solids (s), and aqueous solutions (aq) are denoted in the chemical equations to provide further insight into the reaction conditions. For instance, in \(\mathrm{BaCl}_2(aq) + 2 \mathrm{AgNO}_3(aq) \rightarrow 2 \mathrm{AgCl}(s) + \mathrm{Ba(NO}_3\mathrm{)_2}(aq)\), \(\mathrm{BaCl}_2\) in aqueous solution reacts with silver nitrate \(\mathrm{AgNO}_3\) also in aqueous solution to form solid silver chloride \(\mathrm{AgCl}\) and aqueous barium nitrate \(\mathrm{Ba(NO}_3\mathrm{)_2}\).

Stoichiometric Calculations

The Heart of Quantitative Chemistry

Stoichiometric calculations allow chemists to predict the amounts of reactants needed and products formed in a given reaction. These calculations start with a balanced chemical equation and use the mole concept as their foundation. To perform these calculations, you'll often use conversion factors derived from the coefficients of the balanced equation; these conversion factors tell you the ratio in which reactants combine and the amounts of products formed.

For instance, in the balanced chemical equation \((3) \mathrm{H}_2\mathrm{SO}_4(aq) + (2) \mathrm{Fe}(s) \rightarrow \mathrm{Fe}_2(\mathrm{SO}_4\mathrm{)_3}(aq) + (3) \mathrm{H}_2(g)\), if we start with 0.500 moles of sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)), we can calculate the moles of \(\mathrm{Fe}_2(\mathrm{SO}_4\mathrm{)_3}\) and hydrogen gas (\(\mathrm{H}_2\)) produced by using the mole ratios from the balanced equation. This teaches us that stoichiometry is not only about understanding and balancing equations but also about precisely quantifying the reactants and products within a reaction.