Problem 91 - General Chemistry

Question

How many milliliters of \(0.150 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) are required to react with \(8.20 \mathrm{~g}\) of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), according to the following equation? \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaHCO}_{3}(a q) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) $$

Step-by-Step Solution

Answer

325.3 mL of 0.150 M H2SO4 is needed.

Step 1: Calculate Moles of NaHCO3

First, calculate the number of moles of sodium hydrogen carbonate (NaHCO3) using its given mass and molar mass. The molar mass of NaHCO3 is approximately 84.01 g/mol. Use the formula: \[ \text{Moles of NaHCO}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{8.20 \text{ g}}{84.01 \text{ g/mol}} \approx 0.0976 \text{ moles} \]

Step 2: Use Stoichiometry to Find Moles of H2SO4 Required

From the balanced chemical equation, 1 mole of H2SO4 reacts with 2 moles of NaHCO3. Therefore, the moles of H2SO4 required is half of the moles of NaHCO3 calculated: \[ \text{Moles of } \mathrm{H}_2\mathrm{SO}_4 = \frac{0.0976}{2} \approx 0.0488 \text{ moles} \]

Step 3: Convert Moles of H2SO4 to Volume in Liters

Using the molarity formula, convert the moles of H2SO4 to volume. Recall that Molarity (M) = Moles/Volume (L). Solve for volume: \[ \text{Volume of } \mathrm{H}_2\mathrm{SO}_4 = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.0488 \text{ moles}}{0.150 \text{ M}} = 0.3253 \text{ L} \]

Step 4: Convert Volume to Milliliters

Since 1 liter = 1000 milliliters, multiply the volume in liters by 1000 to convert to milliliters: \[ \text{Volume in mL} = 0.3253 \text{ L} \times 1000 \frac{\text{mL}}{\text{L}} = 325.3 \text{ mL} \]

Key Concepts

Molarity

Molarity, often denoted by the symbol "M," is a crucial concept in chemistry that describes the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Understanding molarity helps chemists quantify how much of a substance is present in a certain volume of liquid.
For instance, in the given exercise, we are dealing with a solution where sulfuric acid, \(\text{H}_2\text{SO}_4\), has a molarity of 0.150 M.
This means there are 0.150 moles of sulfuric acid for every liter of the solution. Knowing this allows us to perform calculations that relate moles of reactants to the volume of solutions used in reactions.

• Molarity is used to convert moles of a solute to volume, using the formula: \( \text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \).
• It guides how solutions react with each other in specific proportions.

Balanced Chemical Equation

A balanced chemical equation ensures that the number of each type of atom is the same on both sides of the equation. This is essential because mass is conserved in chemical reactions, meaning you cannot create or destroy atoms.
In the equation provided in the exercise, sulfuric acid \(\text{H}_2\text{SO}_4\) reacts with sodium hydrogen carbonate \(\text{NaHCO}_3\) to form sodium sulfate \(\text{Na}_2\text{SO}_4\), water \(\text{H}_2\text{O}\), and carbon dioxide \(\text{CO}_2\).
The balanced equation is:\[ \text{H}_2\text{SO}_4(aq) + 2\text{NaHCO}_3(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l) + 2\text{CO}_2(g) \]

• Notice: For every 1 mole of \(\text{H}_2\text{SO}_4\), 2 moles of \(\text{NaHCO}_3\) are required.
• This equation tells us the stoichiometry or the ratio of reactants to products.

Moles Calculation

Calculating moles is a fundamental skill in chemistry, serving as the foundation for stoichiometry. Moles are a measure of how many "entities," such as atoms or molecules, are present. Using moles allows us to quantify and work with chemical reactions.
In the problem, we calculate moles of sodium hydrogen carbonate, \(\text{NaHCO}_3\), using its mass and molar mass:\[ \text{Moles of NaHCO}_3 = \frac{8.20 \text{ g}}{84.01 \text{ g/mol}} = 0.0976 \text{ moles} \]Once we know the moles of \(\text{NaHCO}_3\), we can relate it to the moles of \(\text{H}_2\text{SO}_4\) needed using the balanced equation: \[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{0.0976}{2} = 0.0488 \text{ moles} \]

• Mole calculations allow conversion from mass to moles using: \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).
• We then use mole ratios from balanced chemical equations to find the moles of other reactants or products involved.

Solution Concentration

The concept of solution concentration is integral in chemistry as it provides details about the amount of solute dissolved in a solution. There are different ways to express concentration, but molarity is particularly common. It directly relates to the reactions taking place when solutions are mixed.Using the molarity calculated in the exercise, we have:\[ \text{Volume of } \text{H}_2\text{SO}_4 = \frac{0.0488 \text{ moles}}{0.150 \text{ M}} = 0.3253 \text{ L} \]This shows that to provide the 0.0488 moles of \(\text{H}_2\text{SO}_4\) necessary to react completely with \(8.20 \text{ g}\) of \(\text{NaHCO}_3\), we need 0.3253 liters or 325.3 milliliters of a 0.150 M solution. This calculation hinges on:

• Understanding the definition of molarity in terms of solution volume and moles of solute.
• Being able to convert volume measurements as necessary, often between liters and milliliters.

Solution concentration, therefore, informs interactions between solutes and solvents, thereby dictating the direction and capacity of reactions.