Question

For each of the following unbalanced chemical equations, suppose that exactly \(5.00 \mathrm{g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{S}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\operatorname{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{Mn}\left(\mathrm{SO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{AgNO}_{3}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)

Short answer

For equation (a) S(s) + H2SO4(aq) → SO2(g) + 2H2O(l), the limiting reactant is H2SO4 with expected product masses of 3.27 grams of SO2 and 1.84 grams of H2O. Apply the same six-step process provided for equations (b), (c), and (d).

Step-by-step solution

Balance the Chemical Equation

Starting with the given equation, balance it by adjusting the number of molecules for each compound: S(s) + H2SO4(aq) → SO2(g) + 2H2O(l)

Determine Moles of Reactants

Using the molar mass of S (32.07 g/mol) and H2SO4 (98.08 g/mol), determine the number of moles of S and H2SO4 given 5.00 grams of each reactant: S: \( n_S = \frac{5.00}{32.07} = 0.1559 \: mol \) H2SO4: \( n_{H2SO4} = \frac{5.00}{98.08} = 0.0510 \: mol \)

Calculate Mole Ratios

Determine the mole ratio by dividing the moles of each reactant with their stoichiometric coefficients: S: \( \frac{n_S}{1} = 0.1559 \) H2SO4: \( \frac{n_{H2SO4}}{1} = 0.0510 \)

Identify Limiting Reactant

Since the S/H2SO4 mole ratio (0.0510) is smaller than the S mole ratio (0.1559), H2SO4 is the limiting reactant.

Calculate Moles of Products Formed

Based on the balanced equation, 1 mole of H2SO4 forms 1 mole of SO2 and 2 moles of H2O. Therefore, the moles of products formed are: SO2: \( n_{SO2} = n_{H2SO4} = 0.0510 \: mol \) H2O: \( n_{H2O} = 2 \times n_{H2SO4} = 2 \times 0.0510 = 0.102 \: mol \)

Convert Moles of Products to Grams

Using the molar mass of SO2 (64.07 g/mol) and H2O (18.02 g/mol), calculate the mass of each product expected: SO2: \( m_{SO2} = n_{SO2} \times 64.07 = 0.0510 \times 64.07 = 3.27 \: g \) H2O: \( m_{H2O} = n_{H2O} \times 18.02 = 0.102 \times 18.02 = 1.84 \: g \) The limiting reactant for equation a is H2SO4, and the expected mass of products formed are 3.27 grams of SO2 and 1.84 grams of H2O. Continue with the same six steps for parts b, c, and d of the problem.

Key concepts

Chemical Equation Balancing

Understanding how to balance a chemical equation is the foundation for many concepts in chemistry, including limiting reactant calculations. A balanced chemical equation contains the same number of atoms of each element on both sides of the reaction. Here's a simple guide to balancing:

First, identify the atoms present in the reactants and products. Count the number of atoms for each element and compare the reactants and products. If they are not equal, add stoichiometric coefficients, which are numbers placed in front of compounds in the equation to balance the atom counts. Remember, these coefficients indicate the relative amount of molecules or formula units needed to balance the reaction. It's crucial to never change the subscripts in the chemical formulas as this would alter the substances involved in the reaction.

To practice, try balancing simple reactions before moving on to more complex ones. Once you master this skill, interpreting and predicting chemical reactions becomes much easier.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is a quantitative relationship based on the balanced equations and the conservation of mass. To master stoichiometry, begin by ensuring your chemical equation is balanced. Then, use the stoichiometric coefficients to guide your calculations. These numbers tell you the molar ratio between the reactants and products, which is essential in determining how much of each substance will react or be produced.

Applying Stoichiometry

For instance, when calculating how much product can be formed from a certain amount of reactants, you would use the molar ratios derived from the balanced equation to convert between moles of different chemicals.

In the exercise provided, stoichiometry helps you predict the mass of products formed from a set amount of reactants, providing a direct application of this concept.

Molar Mass

Molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. The molar mass of elements is listed on the periodic table and is usually expressed in grams per mole (g/mol). For compounds, you calculate the molar mass by summing the molar masses of all the atoms in the molecule.

Calculating Molar Mass

As an example, water (H2O) has a molar mass that can be calculated by adding the molar masses of two hydrogen atoms and one oxygen atom. This concept is fundamental in converting grams to moles, which allows for stoichiometric calculations in chemical reactions.

In the exercise, you used the molar mass to determine how many moles 5.00 grams of each reactant correspond to, which is a key step in identifying the limiting reactant and predicting the masses of products formed.

Mole Ratio

The mole ratio is a critical part of stoichiometry that refers to the ratio of moles of one substance to the moles of another substance in a balanced equation. These ratios come directly from the stoichiometric coefficients in a chemical equation.

For instance, in the formation of water from hydrogen and oxygen (H2 + O2 -> H2O), the balanced equation would show a 2:1:2 mole ratio for hydrogen:oxygen:water. This means that every 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Understanding and applying mole ratios allow us to predict the amount of product from a given amount of reactant and vice versa. The exercise solution beautifully demonstrates calculating and using mole ratios to find the limiting reactant and project product formation.

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers in front of molecules in a balanced chemical equation. They indicate the proportion of reactants that combine and the number of products formed. It's these coefficients that determine the mole ratio between different substances in a reaction.

The Importance of Stoichiometric Coefficients

These coefficients are the key to understanding quantities in reactions. In the context of the exercise, we see stoichiometric coefficients being used to equate the moles of reactants to the moles of products, revealing which reactant will limit the reaction and how much of each product will be formed. It's a fine illustration of using stoichiometric coefficients to tie together chemical equation balancing, mole ratios, and molar mass to solve a practical chemistry problem.