Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. The unbalanced chemical equation for this reaction is given below: $$\mathrm{SO}_{2}(g)+\mathrm{NaOH}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(i)$$ Assuming you react 38.3 g sulfur dioxide with 32.8 g sodium hydroxide and assuming that the reaction goes to completion, calculate the mass of each product formed.

The first crucial step in many stoichiometry calculations is balancing chemical equations. This process ensures that the law of conservation of mass is followed, meaning the number of atoms of each element remains the same before and after a chemical reaction. To balance an equation, coefficients are added before the reactant and product formulas until the number of atoms of each element is the same on both sides of the equation.

For instance, the reaction between sulfur dioxide and sodium hydroxide initially reads as: \[\begin{equation}\mathrm{SO}_{2}(g) + \mathrm{NaOH}(s) \longrightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(l)\end{equation}\]Balancing this requires recognizing that while sulfur and oxygen are balanced, the sodium and hydrogen are not. By adding a coefficient of two before NaOH and balancing hydrogen and oxygen accordingly, we achieve the balanced reaction: \[\begin{equation}\mathrm{SO}_{2}(g) + 2\mathrm{NaOH}(s) \longrightarrow \mathrm{Na}_{2}\mathrm{SO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(l)\end{equation}\]This balanced equation is now ready for stoichiometric analysis and has paved the way for identifying the limiting reactant and calculating masses of the products.

Identifying the limiting reactant is essential in predicting the amount of product that can be formed during a chemical reaction. The limiting reactant is the substance that is completely consumed first, limiting the reaction's progression and the amount of product formed. In our example, upon calculating moles of sulfur dioxide and sodium hydroxide, we compare the ratios of reactants to their coefficients in the balanced equation.

When comparing the 0.597 moles of SO2 to the needed 0.410 moles of NaOH (since NaOH is needed in a 2:1 ratio), it becomes clear that NaOH will limit the amount of product formed. With this information, further stoichiometric calculations are based on the amount of the limiting reactant, ensuring accurate predictions of the masses of products generated from the reaction.

Molar mass acts as a bridge between the macroscopic world of grams that we measure in the lab and the microscopic world of molecules and atoms. It is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). This value is crucial for converting between the mass of a substance and the amount in moles, facilitating quantitative analysis in chemistry.

During stoichiometry calculations, we use the molar masses of reactants and products to determine how many moles are present or to find out how many grams of product will be formed. For example, with a molar mass of 126.1 g/mol for sodium sulfite (Na2SO3) and 18.015 g/mol for water (H2O), we can convert the moles of product expected from the limiting reactant to the mass of products that will be formed in the reaction. These conversions serve as a foundation for predicting and measuring the outcomes of chemical reactions in practical scenarios.