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Stoichiometry is the method of calculating the quantitative relationships between the amounts of reactants and products in a chemical reaction. In simpler terms, it helps us understand how much of each chemical is needed or produced. This is a critical step in solving chemistry problems as it allows us to predict the outcomes of reactions precisely.
Understanding the concept of the mole ratio is crucial in stoichiometry. In our exercise, the balanced chemical equation shows that 2 moles of nitric acid \(\text{(HNO}_3)\) are required to react with 1 mole of sodium carbonate \(\text{(Na}_2\text{CO}_3)\).
This specific ratio comes directly from the coefficients of the reactants in the balanced equation. Using this ratio, we can determine how many moles of one reactant are needed given the amount of the other reactant.

• The reaction needs twice as much nitric acid as sodium carbonate.
  
• Understanding this ratio allows us to calculate the amount of each substance necessary for the reaction to proceed without leftovers.
  

By mastering stoichiometry, we can efficiently design reactions to ensure all chemicals are fully used and predictably handle complex chemical processes.

Molarity is a measure of the concentration of a solution, denoted as \(M\), and is defined as the number of moles of solute per liter of solution. In other words, it tells us how much of a substance is in a given volume of liquid.

For example, a \(0.230\,M\) solution of nitric acid means there are 0.230 moles of nitric acid dissolved in every liter of solution.

• Molarity is useful when we want to know how strong or weak a solution is.
• It helps us determine how much of a chemical is involved in a reaction.

In our given problem, we use molarity to find out how many moles of \(Na_2CO_3\) are in \(49.0\,mL\) of solution:
\[\text{Moles of } Na_2CO_3 = 0.330 \, \text{M} \times 0.0490 \, \text{L} = 0.01617 \, \text{mol}\]
We then use the molarity of \(HNO_3\) to find the volume needed for the reaction.
Molarity is essential for solving many types of chemical problems, especially when dealing with solutions.

Balancing chemical equations is necessary because it ensures that the law of conservation of mass is observed, meaning matter is not created or destroyed. This is done by ensuring the same number of each type of atom appears on both sides of the equation. When equations are balanced, they provide the accurate stoichiometry needed for calculations.

The reaction given in the problem:\[2 \text{HNO}_{3}(aq) + \text{Na}_{2} \text{CO}_{3}(aq) \rightarrow 2 \text{NaNO}_{3}(aq) + \text{H}_{2}\text{O}(l) + \text{CO}_{2}(g)\]is already balanced. Here’s why balancing is important:

• Ensures that we have the correct amounts of reactants producing the intended products.
• Facilitates the use of stoichiometry to gauge reactant needs and product amounts.

In this balanced equation, 2 molecules of \(\text{HNO}_3\) react with 1 molecule of \(\text{Na}_2\text{CO}_3\), producing 2 molecules of \(\text{NaNO}_3,\) and one molecule each of \(\text{H}_2\text{O}\) and \(\text{CO}_2\).
Balancing equations is a vital skill in chemistry, serving as the foundation for understanding and conducting chemical reactions.