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Understanding chemical reactions is crucial when solving stoichiometry problems, like the decomposition of sodium nitrate. A chemical reaction involves the transformation of one or more substances into different substances. This happens when atoms rearrange, forming new bonds. Chemical reactions can be represented by balanced chemical equations.

In this particular example, we have sodium nitrate (\(\mathrm{NaNO}_3\)) decomposing into sodium nitrite (\(\mathrm{NaNO}_2\)) and oxygen gas (\(\mathrm{O}_2\)). To solve this kind of problem, first write out the balanced equation: \[2 \mathrm{NaNO}_3 (s) \rightarrow 2 \mathrm{NaNO}_2 (s) + \mathrm{O}_2 (g)\]
This equation shows that two units of sodium nitrate produce two units of sodium nitrite and one of oxygen gas. Balancing the chemical equation is essential as it ensures that mass is conserved and that the number of each type of atom is equal on both sides of the equation.

Properly understanding chemical reactions and their equations is necessary for effectively calculating the amounts of reactants and products, ensuring you get accurate stoichiometry results.

Molar mass is a key concept in chemical calculations. It represents the mass of one mole of a given substance and is expressed in units of grams per mole (g/mol). Knowing the molar mass allows us to convert between the mass of a substance and the amount of substance in moles.

To determine the molar mass of a compound, you sum up the atomic masses of all the atoms present in its molecular formula. For sodium nitrite (\(\mathrm{NaNO}_2\)), its molar mass is calculated by adding the atomic masses of sodium (Na), nitrogen (N), and oxygen (O). That is \(22.99\ \text{g/mol (Na)} + 14.01\ \text{g/mol (N)} + 16.00 \ \text{g/mol (O)} \times 2 = 69.00\ \text{g/mol}\).

The understanding of molar mass allows us to move between amounts in grams and moles, which forms the basis for stoichiometric calculations in chemical reactions.

Chemical equations are symbolic representations of chemical reactions. They provide information about the reactants and products involved, as well as their relative proportions. This is evident through the coefficients, which are the numbers placed before compounds.

The task here involves the decomposition of sodium nitrate, represented by \[2 \mathrm{NaNO}_3 (s) \rightarrow 2 \mathrm{NaNO}_2 (s) + \mathrm{O}_2 (g)\].This equation is balanced, meaning both sides have equal numbers of each type of atom, maintaining the conservation of mass. The coefficients also give us a ratio of moles: 2 moles of sodium nitrate produce 2 moles of sodium nitrite and 1 mole of oxygen gas.

Balancing a chemical equation is crucial because it ensures our calculations in a stoichiometric problem are accurate. By understanding the role of chemical equations, students can more easily determine how much reactant is needed or how much product can be made.

Percent composition tells us the percentage by mass of each element in a compound or the proportion of a chemical within a mixture. In the problem given, we determine the percentage of sodium nitrate in an impure sample by finding its mass contribution to the total sample.

To find percent composition, divide the mass of the component (sodium nitrate) by the total mass of the sample and multiply by 100. For example, if the mass of sodium nitrate is \(0.288\ \mathrm{g}\) and the mass of the original sample is \(0.423\ \mathrm{g}\), the percent composition is calculated as follows:\[\text{Percent of sodium nitrate} = \frac{0.288\ \mathrm{g}}{0.423\ \mathrm{g}} \times 100\% \approx 68.1\%\].

This shows that the original sample contains approximately 68.1% sodium nitrate by mass. Understanding percent composition is beneficial in fields such as chemical analysis and material characterization.