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 \(0.125 \mathrm{M} \mathrm{HNO}_{3} ?\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HNO}_{3}(\mathrm{aq}) \longrightarrow\) $$2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$

Step-by-Step Solution

Answer

0.331 g of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is required.

Step 1: Calculate Moles of \(\mathrm{HNO}_{3}\)

First, we need to find the number of moles of \(\mathrm{HNO}_{3}\) present. Use the formula for moles, which is \(n = M \times V\), where \(M\) is the molarity and \(V\) is the volume in liters. Convert 50.0 mL to liters by dividing by 1000, resulting in 0.050 L. Then calculate:\[n = (0.125 \ \mathrm{M}) \times (0.050 \ \mathrm{L}) = 0.00625 \ \mathrm{mol}\mathrm{es}\ \mathrm{of}\ \mathrm{HNO}_{3}\]

Step 2: Use Stoichiometry to Find Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) Needed

According to the balanced chemical equation, 2 moles of \(\mathrm{HNO}_{3}\) react with 1 mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). Use the stoichiometric relationship:\[\text{Moles of } \mathrm{Na}_{2}\mathrm{CO}_{3} = \frac{0.00625\ \mathrm{mol}\ \mathrm{HNO}_{3}}{2} = 0.003125 \ \mathrm{moles}\ \mathrm{Na}_{2}\mathrm{CO}_{3}\]

Step 3: Calculate the Mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\)

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 \[M(\mathrm{Na}_{2}\mathrm{CO}_{3}) = 2(23.0) + 12.0 + 3(16.0) = 106.0 \ \mathrm{g/mol}\]Multiply the moles by the molar mass to find the mass:\[\text{Mass of } \mathrm{Na}_{2}\mathrm{CO}_{3} = 0.003125 \ \mathrm{moles} \times 106.0 \ \mathrm{g/mol} = 0.33125 \ \mathrm{g}\]

Step 4: Conclude the Calculation

Round the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) to the appropriate number of significant figures, which in this case, is three significant figures given by 0.125 M and 50 mL, thus:\[\text{Mass of } \mathrm{Na}_{2}\mathrm{CO}_{3} \approx 0.331 \ \mathrm{g}\]

Key Concepts

Chemical Reactions

In chemistry, a chemical reaction occurs when substances, known as reactants, are transformed into different substances, called products. In the given exercise, the substances undergoing the transformation are \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HNO}_{3}\). The products formed include \(\mathrm{NaNO}_{3}\), \(\mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\). Understanding chemical reactions is crucial for predicting the amounts of reactants needed and products formed. This particular reaction can be described by the balanced equation:\[\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) + 2 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{NaNO}_{3}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell)\]This balanced equation tells us that 1 mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) reacts with 2 moles of \(\mathrm{HNO}_{3}\) to form the products, demonstrating the stoichiometric coefficients. This relationship is critical for calculations involving the quantity of reactants or products in a chemical reaction.

Molarity

Molarity is a key concept in chemistry that refers to the concentration of a solution. It is defined as the number of moles of solute (in this case, \(\mathrm{HNO}_{3}\)) per liter of solution. The formula for molarity is:\[M = \frac{n}{V}\]Where \(M\) is the molarity, \(n\) is the number of moles, and \(V\) is the volume in liters. For the exercise, a 0.125 M (molar) solution of \(\mathrm{HNO}_{3}\) was used. We had 50.0 mL of this acid solution, which is equal to 0.050 liters since there are 1000 milliliters in a liter. By using these values, we can calculate the moles of \(\mathrm{HNO}_{3}\) in the solution, which is necessary for moving forward with stoichiometric calculations.

Moles Calculation

Calculating the number of moles of a reactant or product is a fundamental part of stoichiometry. In this specific situation, we need to calculate the moles of \(\mathrm{HNO}_{3}\) that are present in the solution to determine how much \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is required for the complete reaction. We use the formula:\[n = M \times V\]Given that the molarity \(M\) is 0.125 M and the volume \(V\) is 0.050 L, we compute:\[n = 0.125\, \mathrm{mol/L} \times 0.050\, \mathrm{L} = 0.00625\, \mathrm{mol}\]This calculation tells us that there are 0.00625 moles of \(\mathrm{HNO}_{3}\) in the solution. According to the balanced chemical equation, 2 moles of \(\mathrm{HNO}_{3}\) are required for every mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), which helps us determine the required moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).

Molar Mass

Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It serves as a conversion factor between moles and grams, allowing chemists to calculate the amount of substance present in a sample. To calculate molar mass, sum up the atomic masses of all atoms present in the compound based on the chemical formula. For \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), the calculation is as follows:- Sodium (Na): 23.0 g/mol; 2 atoms contribute \(2 \times 23.0 = 46.0\, \mathrm{g/mol}\)- Carbon (C): 12.0 g/mol; 1 atom contributes \(1 \times 12.0 = 12.0\, \mathrm{g/mol}\)- Oxygen (O): 16.0 g/mol; 3 atoms contribute \(3 \times 16.0 = 48.0\, \mathrm{g/mol}\)Adding these gives us the molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\):\[46.0 + 12.0 + 48.0 = 106.0\, \mathrm{g/mol}\]Using the molar mass, you can convert moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) into grams, which is necessary for finding the solution to the problem."