Question

Consider the reaction represented by the chemical equation $$\mathrm{KOH}(s)+\mathrm{SO}_{2}(g) \rightarrow \mathrm{KHSO}_{3}(s)$$ since the coefficients of the balanced chemical equation are all equal to \(1,\) we know that exactly \(1 \mathrm{g}\) of KOH will react with exactly \(1 \mathrm{g}\) of \(\mathrm{SO}_{2}\). True or false? Explain.

Short answer

False. 1g of KOH will not react exactly with 1g of SO2. Based on the stoichiometry and molar masses of the reactants, 56g of KOH will fully react with 64g of SO2 in order to maintain the mole-to-mole ratio in the balanced chemical equation.

Step-by-step solution

Write down the balanced chemical equation

First, we need to write down the balanced chemical equation: \( \mathrm{KOH}(s)+\mathrm{SO}_{2}(g) \rightarrow \mathrm{KHSO}_{3}(s) \) The coefficients in front of each species indicate the number of moles of each reactant required for the reaction to proceed.

Calculate the molar masses of the reactants

Next, we need to determine the molar masses of the reactants KOH and SO2. This information will help us to calculate the mass ratios of these reactants in the reaction. Molar mass of KOH = \( \mathrm{(K: 39) + (O: 16) + (H: 1)} = 56 g/mol \) Molar mass of SO2 = \( \mathrm{(S: 32) + (O: 16) * 2 } = 64 g/mol \)

Convert grams of reactants to moles and compare

Now, we need to convert the 1g of each reactant we are given into moles, using their respective molar masses. Moles of KOH = \( \dfrac{1g}{56 g/mol} = \dfrac{1}{56} mol \) Moles of SO2 = \( \dfrac{1g}{64 g/mol} = \dfrac{1}{64} mol \) From the balanced equation, it is clear that the stoichiometry is 1:1, meaning that 1 mole of KOH needs to react with 1 mole of SO2.

Determine reactant quantities and provide an explanation

Now we compare the moles of each reactant to the stoichiometric requirements of the balanced chemical equation: Since \( \dfrac{1}{56} mol \) of KOH reacts with \( \dfrac{1}{56} mol \) of SO2, this situation is less than 1 mole of KOH (and SO2). In conclusion, the statement in the exercise is false. 1g of KOH will not react exactly with 1g of SO2. Instead, based on the stoichiometry and molar masses of the reactants, 56g of KOH will fully react with 64g of SO2 in order to maintain the mole-to-mole ratio in the balanced chemical equation.

Key concepts

Balanced Chemical Equation

Understanding a balanced chemical equation is crucial when studying chemistry, as it implies that the number of atoms for each element involved remains the same before and after a chemical reaction. This is in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed in a chemical reaction.

For instance, in the chemical equation \( \mathrm{KOH}(s) + \mathrm{SO}_{2}(g) \rightarrow \mathrm{KHSO}_{3}(s) \), each element is accounted for on both the reactant and product side of the equation. Here, the coefficients indicate the mole-to-mole ratio in which substances react and are produced. In this case, all coefficients are one, suggesting that one mole of potassium hydroxide reacts with one mole of sulfur dioxide to produce one mole of potassium bisulfite.

It’s a common misconception that these stoichiometric coefficients suggest a weight-to-weight relationship, when in reality, they represent a mole-to-mole ratio. Thus, it is incorrect to assume that one gram of \(\mathrm{KOH}\) will react with one gram of \(\mathrm{SO}_{2}\) just because their coefficients are one. To truly understand the reaction quantities, you must delve deeper into the molar masses of the substances involved.

Molar Mass

Molar mass is a fundamental concept in chemistry, providing a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. It is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

The molar mass of a compound can be calculated by summing the individual molar masses of the elements that compose it, according to their respective elemental molar masses and the number of atoms of each element within the chemical formula. For example, the molar mass of potassium hydroxide (\(\mathrm{KOH}\)) can be calculated as follows: \( \mathrm{(K: 39) + (O: 16) + (H: 1)} = 56 \ g/mol \). Similarly, for sulfur dioxide (\(\mathrm{SO}_{2}\)), the calculation is \( \mathrm{(S: 32) + (O: 16) \times 2} = 64 \ g/mol \).

Knowing the molar mass is essential for converting between grams of a substance and moles, which allows you to use the balanced chemical equation to predict how much of a reactant you need or how much of a product you can expect from a given mass of reactants.

Mole-to-Mole Ratio

The mole-to-mole ratio is derived from the balanced chemical equation and is a key concept in stoichiometry. It tells us how many moles of one chemical react with moles of another chemical. In the equation \( \mathrm{KOH} + \mathrm{SO}_{2} \rightarrow \mathrm{KHSO}_{3} \), the stoichiometry implies a 1:1 mole ratio between \(\mathrm{KOH}\) and \(\mathrm{SO}_{2}\). To translate this conceptual ratio into a practical laboratory setting, it is necessary to utilize the molar masses of the reactants to calculate the number of moles present in a given mass of each reactant.

Using the provided exercise, we learned that one gram of each substance does not provide equimolar amounts due to their differing molar masses: \( 1g \mathrm{KOH} = \frac{1}{56} \, \mathrm{mol} \) and \( 1g \mathrm{SO}_{2} = \frac{1}{64} \, \mathrm{mol} \). Clearly, these are not equivalent and do not satisfy the 1:1 ratio required by the equation. It is crucial to understand that mass does not directly equate to the number of moles, as the molar mass must be taken into account to find the correct mole-to-mole ratio for the reactants.