Problem 62 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

0.331 grams of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) are needed.

Step 1: Determine moles of HNO3

First, calculate the moles of \( \mathrm{HNO}_{3} \) in the given solution using the formula: \( \text{{moles}} = \text{{volume}} \times \text{{molarity}} \). The volume should be in liters, hence convert \( 50.0 \ \mathrm{mL} \) to liters by dividing by 1000, resulting in \( 0.0500 \ \mathrm{L} \). Thus, the moles of \( \mathrm{HNO}_{3} \) are: \[ \text{{moles}} = 0.0500 \ \mathrm{L} \times 0.125 \ \mathrm{M} = 0.00625 \ \text{{moles}}. \]

Step 2: Determine moles of Na2CO3 needed

From the balanced chemical equation, 1 mole of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) reacts with 2 moles of \( \mathrm{HNO}_{3} \). Using the stoichiometry of the reaction, calculate the moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) needed: \[ \text{{moles of }} \mathrm{Na}_{2} \mathrm{CO}_{3} = \frac{0.00625 \ \text{{moles of }} \mathrm{HNO}_{3}}{2} = 0.003125 \ \text{{moles}}. \]

Step 3: Calculate mass of Na2CO3 required

Now, calculate the mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) required using its molar mass. The molar mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) is calculated as follows: Na (22.99) \( \times 2 \), C (12.01) \( \times 1 \), and O (16.00) \( \times 3 \). Adding these gives \( 105.99 \ \mathrm{g/mol} \). Then, calculate the mass: \[ \text{{mass}} = 0.003125 \ \text{{moles}} \times 105.99 \ \mathrm{g/mol} = 0.331 \ \mathrm{g}. \]

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It's defined as the number of moles of a solute dissolved per liter of solution.

• Formula: Molarity (M) = Moles of solute / Volume of solution in liters
• A solution with a molarity of 1 M has one mole of solute in every liter.

To solve chemistry problems involving molarity, it's crucial to convert volumes to liters. In the given exercise, we converted 50.0 mL of HNO₃ to liters by dividing by 1000, resulting in 0.0500 L. Then, we calculated the moles of HNO₃ using the formula \( \text{moles} = 0.0500 \, \text{L} \times 0.125 \, \text{M} = 0.00625 \, \text{moles}\). This calculation helps determine the amount of reactants and products in a reaction, setting the stage for further stoichiometry calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They are represented by balanced chemical equations, where the number of atoms for each element is the same on both sides.In our exercise, the balanced equation is: \( \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) + 2 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{NaNO}_{3}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell) \).

• This tells us 1 mole of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) reacts with 2 moles of \( \mathrm{HNO}_{3} \) to produce 2 moles of \( \mathrm{NaNO}_{3} \).

The stoichiometry of the reaction allows us to calculate how much \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) is needed. Because each mole of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) requires 2 moles of \( \mathrm{HNO}_{3} \), knowing the moles of \( \mathrm{HNO}_{3} \) helps us find the required moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) through the ratio \( \frac{1}{2} \), resulting in 0.003125 moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \).

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (\(\, \mathrm{g/mol}\)). It is calculated by adding together the atomic masses of the elements in a compound.For \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), the molar mass is determined by combining the atomic masses:

• Sodium (\( \mathrm{Na} \)): 22.99 \(\, \mathrm{g/mol}\), two atoms contribute \( 2 \times 22.99 = 45.98 \).
• Carbon (\( \mathrm{C} \)): 12.01 \(\, \mathrm{g/mol}\), one atom contributes \( 12.01 \).
• Oxygen (\( \mathrm{O} \)): 16.00 \(\, \mathrm{g/mol}\), three atoms contribute \( 3 \times 16.00 = 48.00 \).

Adding these values gives \( 105.99 \ \, \mathrm{g/mol} \).Using the molar mass, convert moles of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) to mass:\[ \text{mass} = 0.003125 \text{ moles} \times 105.99 \ \, \mathrm{g/mol} = 0.331 \ \, \mathrm{g} \].Knowing how to calculate molar mass is essential for finding the total mass of a compound in chemical reactions, providing a crucial step for stoichiometric calculations.