How many grams of \(\mathrm{NaNO}_{2}\) form when 256 \(\mathrm{g} \mathrm{NaNO}_{3}\) react? The yield is 91\(\% .\) $$2 \mathrm{NaNO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s)+\mathrm{O}_{2}(g)$$

Understanding chemical equation balancing is crucial for studying chemical reactions. It is based on the Law of Conservation of Mass, which states that in a chemical reaction, the mass of the reactants equals the mass of the products. To balance a chemical equation, every element must have the same number of atoms on both the reactant and product sides of the equation.

Take the textbook exercise as an example, where the equation is balanced as follows:
\[2 \mathrm{NaNO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{2}(s) + \mathrm{O}_{2}(g)\]
Here, we see that for every two moles of sodium nitrate (\(\mathrm{NaNO}_{3}\)), two moles of sodium nitrite (\(\mathrm{NaNO}_{2}\)) and one mole of oxygen gas (\(\mathrm{O}_{2}\)) are produced. This balanced equation is essential for stoichiometric calculations because it informs us of the proportion in which reactants combine and the products form.

The molar mass of a substance is the weight of one mole of that substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms present in the molecular formula of the substance.

For example, in our exercise, the molar mass of sodium nitrate (\(\mathrm{NaNO}_{3}\)) is calculated as follows:
\[23 (\text{{for Na}}) + 14 (\text{{for N}}) + 3 \times 16 (\text{{for each O}}) = 85 \text{{ g/mol}}\]
Knowing the molar mass allows us to convert between grams and moles, which is a pivotal step in stoichiometric calculations. It's the bridge that connects the mass of a substance with the amount in moles, enabling quantitative analysis of a chemical reaction.

Stoichiometry is the quantitative aspect of chemistry that deals with the amounts of reactants and products in a chemical reaction. It involves calculations that use the balanced chemical equation to determine the mass, moles, and number of particles of substances involved.

In the given exercise, we begin with the provided mass of sodium nitrate (256 g of \(\mathrm{NaNO}_{3}\)) and use stoichiometry to find out how much sodium nitrite (\(\mathrm{NaNO}_{2}\)) forms. Stoichiometric steps include:

• Determining the mole ratio from the balanced equation
• Calculating the amount of moles from given mass
• Applying the actual yield percentage to find the effective product
• Converting moles back to grams

Following these steps systematically will lead to the accurate determination of the desired quantity, in this case, 188 grams of \(\mathrm{NaNO}_{2}\) with a 91% yield. Stoichiometry is a fundamental technique for predicting the outcomes of reactions and for scaling them up or down in practical applications.