Question

What mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) in grams, is required for complete reaction with \(50.0 \mathrm{mL}\) of \(0.125 \mathrm{M} \mathrm{HNO}_{3} ?\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$

Short answer

0.331 grams of \( \mathrm{Na}_2 \mathrm{CO}_3 \) are needed.

Step-by-step solution

Write the Balanced Chemical Equation

The balanced equation already provided is as follows: \( \mathrm{Na}_2 \mathrm{CO}_3(\mathrm{aq}) + 2 \mathrm{HNO}_3(\mathrm{aq}) \rightarrow 2 \mathrm{NaNO}_3(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\ell) \). This equation shows that one mole of sodium carbonate reacts with two moles of nitric acid.

Calculate the Moles of Nitric Acid

First, we find the moles of \( \mathrm{HNO}_3 \) using its molarity and the volume provided. Use the formula: \( \text{moles} = \text{molarity} \times \text{volume (in liters)} \). So, \( 0.125 \, \mathrm{mol/L} \times 0.0500 \, \mathrm{L} = 0.00625 \, \mathrm{mol} \).

Determine the Moles of Sodium Carbonate Required

From the balanced equation, 1 mole of \( \mathrm{Na}_2 \mathrm{CO}_3 \) reacts with 2 moles of \( \mathrm{HNO}_3 \). Therefore, the moles of \( \mathrm{Na}_2 \mathrm{CO}_3 \) required are \( \frac{0.00625}{2} = 0.003125 \) moles.

Calculate the Mass of Sodium Carbonate Required

To find the mass, use the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \). The molar mass of \( \mathrm{Na}_2 \mathrm{CO}_3 \) is \( 2 \times 22.99 \, (\mathrm{Na}) + 12.01 \, (\mathrm{C}) + 3 \times 16.00 \, (\mathrm{O}) = 105.99 \, \mathrm{g/mol} \). So, \( 0.003125 \, \mathrm{mol} \times 105.99 \, \mathrm{g/mol} = 0.331 \, \mathrm{g} \).

Key concepts

Balanced Chemical Equation

In chemistry, a balanced chemical equation is pivotal to understanding how reactants convert to products. It provides a concise way to represent a chemical reaction. In a balanced equation, the number of atoms for each element involved is the same on both sides of the reaction. This aligns with the law of conservation of mass, which states that mass is neither created nor destroyed during a chemical reaction.

For the reaction between sodium carbonate (\( \mathrm{Na}_2 \mathrm{CO}_3 \)) and nitric acid (\( \mathrm{HNO}_3 \)), the balanced equation shows the stoichiometric relationship:

• \(1\) mole of \( \mathrm{Na}_2 \mathrm{CO}_3 \)
• reacts with \(2\) moles of \( \mathrm{HNO}_3 \)
• to produce \(2\) moles of \( \mathrm{NaNO}_3 \), \(1\) mole of \( \mathrm{CO}_2 \), and \(1\) mole of \( \mathrm{H}_2 \mathrm{O} \).

This balanced equation enables us to predict the amounts of reactants needed and products formed in a reaction.

Ensuring a chemical equation is balanced is crucial to performing precise calculations in stoichiometry, which involves determining the quantitative relationships between reactants and products in a chemical reaction.

Molarity

Molarity is a measure of the concentration of a solution and is defined as the number of moles of a solute per liter of solution. It is expressed in moles per liter (mol/L), often abbreviated as "M".

For the reaction, we needed to calculate the moles of \( \mathrm{HNO}_3 \) using its molarity. The problem provided a molarity of \(0.125 \mathrm{M}\) and a volume of \(50.0 \mathrm{mL}\) (which must be converted to liters by dividing by 1000):

\(\text{Moles of } \mathrm{HNO}_3 = 0.125 \mathrm{M} \times 0.0500 \mathrm{L} = 0.00625 \text{ moles}\).

This calculation shows the direct relationship between molarity, volume, and moles of a solute, providing a clear method for determining quantities in a chemical reaction. Understanding molarity is vital for mixing solutions accurately in laboratory settings, and for calculating concentrations when reactions occur in solutions.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is determined by summing the atomic masses of all atoms in a molecular formula.

In the exercise, the molar mass of sodium carbonate \( \mathrm{Na}_2 \mathrm{CO}_3 \) was calculated as follows:

• Sodium (\( \mathrm{Na} \)): \(22.99 \times 2 = 45.98 \) g/mol
• Carbon (\( \mathrm{C} \)): \(12.01 \) g/mol
• Oxygen (\( \mathrm{O} \)): \(16.00 \times 3 = 48.00 \) g/mol
• Total molar mass: \(45.98 + 12.01 + 48.00 = 105.99 \) g/mol

Once you have the molar mass, you can calculate the mass of a substance in a given number of moles using the formula:

\(\text{mass} = \text{moles} \times \text{molar mass}\).

In this case, with \(0.003125\) moles of \( \mathrm{Na}_2 \mathrm{CO}_3 \), the mass needed is \(0.003125 \times 105.99 = 0.331 \) grams. Knowledge of molar mass is essential for converting between mass and moles, which is a common requirement in stoichiometry.