Calculating Sulfuric Acid Needed To React With Sodium Hydroxide

Introduction

In this article, we will delve into a stoichiometry problem involving the reaction between sulfuric acid ($H_2SO_4$) and sodium hydroxide ($NaOH$). This is a classic example of an acid-base neutralization reaction, a fundamental concept in chemistry. Our primary goal is to determine the precise amount of sulfuric acid, measured in grams, required to completely react with a given quantity of sodium hydroxide, specifically 79.3 grams. This calculation is crucial in various chemical applications, including titrations, industrial processes, and laboratory experiments. Understanding the mole ratios and molar masses of the reactants is paramount to accurately solving this problem. We will meticulously walk through the steps, ensuring clarity and comprehension of each stage. So, let's embark on this chemical journey to unravel the quantitative aspects of this important reaction. This involves converting the mass of sodium hydroxide to moles, using the balanced chemical equation to find the moles of sulfuric acid needed, and finally converting that to grams. By understanding these steps, you can solve a variety of stoichiometry problems.

The Balanced Chemical Equation

The first crucial step in any stoichiometry problem is to have a balanced chemical equation. This equation provides the mole ratios of the reactants and products involved in the reaction. In this case, the balanced equation for the reaction between sodium hydroxide ($NaOH$) and sulfuric acid ($H_2SO_4$) is:

$2 NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 H_2O$

This equation tells us that two moles of sodium hydroxide react with one mole of sulfuric acid to produce one mole of sodium sulfate ($Na_2SO_4$) and two moles of water ($H_2O$). This 2:1 mole ratio between $NaOH$ and $H_2SO_4$ is the cornerstone of our calculation. A balanced equation ensures that the law of conservation of mass is adhered to, meaning that the number of atoms of each element is the same on both sides of the equation. If the equation is not balanced, the mole ratios will be incorrect, leading to inaccurate results. Therefore, always double-check the balancing of your chemical equation before proceeding with any calculations. The coefficients in front of the chemical formulas represent the number of moles of each substance involved in the reaction, and they are essential for determining the stoichiometry of the reaction.

Determining Molar Masses

Before we can perform any calculations, we need to determine the molar masses of the reactants involved: sodium hydroxide ($NaOH$) and sulfuric acid ($H_2SO_4$). The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). To calculate the molar mass, we sum the atomic masses of all the atoms in the chemical formula, which can be found on the periodic table.

For sodium hydroxide ($NaOH$):

• Sodium (Na): 22.99 g/mol
• Oxygen (O): 16.00 g/mol
• Hydrogen (H): 1.01 g/mol

Molar mass of $NaOH$ = 22.99 + 16.00 + 1.01 = 40.00 g/mol

For sulfuric acid ($H_2SO_4$):

• Hydrogen (H): 1.01 g/mol (x2 = 2.02 g/mol)
• Sulfur (S): 32.07 g/mol
• Oxygen (O): 16.00 g/mol (x4 = 64.00 g/mol)

Molar mass of $H_2SO_4$ = 2.02 + 32.07 + 64.00 = 98.09 g/mol

These molar masses are essential conversion factors that allow us to move between grams and moles, which is a crucial step in stoichiometry calculations. Knowing the molar mass of a substance allows us to convert a given mass of that substance into the corresponding number of moles, and vice versa. This conversion is necessary because chemical reactions occur at the molar level, as dictated by the balanced chemical equation. Therefore, accurate molar masses are fundamental to obtaining correct results in stoichiometry problems.

Converting Grams of NaOH to Moles

The next step in our calculation is to convert the given mass of sodium hydroxide ($NaOH$), which is 79.3 grams, into moles. To do this, we will use the molar mass of $NaOH$ that we calculated earlier (40.00 g/mol). The conversion formula is:

Moles = Mass / Molar Mass

So, for $NaOH$:

Moles of $NaOH$ = 79.3 g / 40.00 g/mol = 1.9825 moles

This calculation tells us that 79.3 grams of $NaOH$ is equivalent to 1.9825 moles. Converting to moles is a crucial step because the balanced chemical equation expresses the reaction stoichiometry in terms of mole ratios, not mass ratios. This conversion bridges the gap between the macroscopic world of grams, which we can measure in the laboratory, and the microscopic world of moles, which governs chemical reactions. By converting the mass of $NaOH$ to moles, we can now use the balanced chemical equation to determine the corresponding number of moles of $H_2SO_4$ required for the reaction. This step is essential for accurately predicting the amount of reactants needed or products formed in a chemical reaction. Rounding should be done at the end of the calculation to avoid intermediate errors.

Using the Mole Ratio to Find Moles of $H_2SO_4$

Now that we know the number of moles of $NaOH$ (1.9825 moles), we can use the mole ratio from the balanced chemical equation to determine the number of moles of sulfuric acid ($H_2SO_4$) required. From the balanced equation:

$2 NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 H_2O$

We see that 2 moles of $NaOH$ react with 1 mole of $H_2SO_4$. This gives us a mole ratio of 1 mole $H_2SO_4$ / 2 moles $NaOH$. We can use this ratio to convert the moles of $NaOH$ to moles of $H_2SO_4$:

Moles of $H_2SO_4$ = Moles of $NaOH$ × (1 mole $H_2SO_4$ / 2 moles $NaOH$)

Moles of $H_2SO_4$ = 1.9825 moles $NaOH$ × (1 mole $H_2SO_4$ / 2 moles $NaOH$) = 0.99125 moles

This calculation shows that 0.99125 moles of sulfuric acid are needed to react completely with 1.9825 moles of sodium hydroxide. The mole ratio acts as a conversion factor, allowing us to move from the amount of one substance in the reaction to the amount of another. This is a fundamental concept in stoichiometry, as it enables us to predict the quantities of reactants and products involved in a chemical reaction. Understanding and correctly applying mole ratios is essential for solving a wide range of stoichiometry problems, from simple calculations to more complex multi-step problems.

Converting Moles of $H_2SO_4$ to Grams

The final step is to convert the moles of sulfuric acid ($H_2SO_4$) we calculated (0.99125 moles) back to grams. We will use the molar mass of $H_2SO_4$ (98.09 g/mol) for this conversion. The conversion formula is:

Mass = Moles × Molar Mass

So, for $H_2SO_4$:

Mass of $H_2SO_4$ = 0.99125 moles × 98.09 g/mol = 97.234 grams

Therefore, 97.234 grams of sulfuric acid are needed to react completely with 79.3 grams of sodium hydroxide. This final conversion brings us back to the unit of mass (grams), which is often the most practical unit for measuring amounts of substances in the laboratory or in industrial settings. By converting moles back to grams, we can directly relate the calculated amount to the actual mass of the substance needed for the reaction. This step completes the stoichiometric calculation, providing us with the answer to the original problem.

Final Answer

In conclusion, to react completely with 79.3 grams of sodium hydroxide ($NaOH$), you need 97.234 grams of sulfuric acid ($H_2SO_4$). This calculation involved several key steps: balancing the chemical equation, determining molar masses, converting grams to moles, using the mole ratio from the balanced equation, and converting moles back to grams. Each of these steps is crucial for solving stoichiometry problems accurately. Understanding these principles allows for precise calculations in chemical reactions, which is essential in various fields such as chemistry, chemical engineering, and related sciences. By mastering these concepts, one can confidently approach a wide range of stoichiometric calculations, ensuring accurate and reliable results in chemical experiments and industrial processes. Stoichiometry is a foundational aspect of chemistry, and proficiency in these calculations is vital for success in the field.