Calculating Lead(II) Nitrate Required To Produce 425 Grams Of Sodium Nitrate

Introduction

In the realm of chemistry, understanding stoichiometry is crucial for accurately predicting the amounts of reactants and products involved in a chemical reaction. Stoichiometry is the study of the quantitative relationships or ratios between two or more substances when they undergo a physical change or chemical reaction. This article delves into a specific stoichiometric calculation, focusing on determining the mass of lead(II) nitrate ($Pb(NO_3)_2$) needed to produce 425 grams of sodium nitrate ($NaNO_3$) in a given reaction. To solve this problem, we will meticulously walk through the necessary steps, encompassing balancing the chemical equation, calculating molar masses, and applying stoichiometric principles. This detailed explanation will not only provide the answer but also enhance your comprehension of stoichiometric calculations. This understanding is fundamental not only in academic settings but also in practical applications such as industrial chemistry, pharmaceutical research, and environmental science, where precise measurements and conversions are indispensable. By mastering these concepts, one can efficiently design experiments, optimize chemical processes, and ensure the accuracy of chemical analyses. Stoichiometry is a cornerstone of chemical education and practice, enabling chemists and scientists to predict, control, and manipulate chemical reactions with confidence and precision. The ability to perform stoichiometric calculations accurately is an invaluable skill that underpins many aspects of chemistry and related fields.

Problem Statement

The question we aim to address is: In the following reaction, how many grams of lead(II) nitrate ($Pb(NO_3)_2$) will produce 425 grams of sodium nitrate ($NaNO_3$)? The reaction is:

$Pb(NO_3)_{2(aq)} + 2NaBr_{(aq)} \rightarrow PbBr_{2(s)} + 2NaNO_{3(aq)}$

This problem is a classic example of a stoichiometry question, where we need to use the balanced chemical equation to relate the amount of one substance to another. Stoichiometry, as mentioned earlier, is a crucial concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. The balanced chemical equation provides the mole ratios necessary for these calculations. In this particular scenario, we are given the mass of sodium nitrate produced and are asked to find the mass of lead(II) nitrate required. This requires converting grams to moles, using the stoichiometric ratio from the balanced equation, and then converting back to grams. The challenge lies in accurately performing these conversions and applying the mole ratio correctly. Errors in any of these steps can lead to a significantly different answer. Therefore, a methodical approach, paying close attention to units and significant figures, is essential. Understanding the underlying principles of stoichiometry not only helps in solving this specific problem but also equips one to tackle a wide range of chemical calculations. The ability to relate masses and moles of different substances within a reaction is a fundamental skill for anyone studying or working in chemistry.

Solution

Step 1: Verify the Balanced Chemical Equation

The given chemical equation is:

$Pb(NO_3)_{2(aq)} + 2NaBr_{(aq)} \rightarrow PbBr_{2(s)} + 2NaNO_{3(aq)}$

First, it's essential to ensure that the chemical equation is balanced. A balanced equation follows the law of conservation of mass, stating that matter cannot be created or destroyed in a chemical reaction. This means the number of atoms of each element must be the same on both sides of the equation. By inspecting the equation, we can verify that it is indeed balanced:

• Lead (Pb): 1 atom on both sides
• Nitrate ($NO_3$): 2 units on both sides
• Sodium (Na): 2 atoms on both sides
• Bromine (Br): 2 atoms on both sides

Since the number of atoms for each element is the same on both sides, the equation is balanced. This balanced equation is the foundation for our stoichiometric calculations because it provides the mole ratios between the reactants and products. For instance, it tells us that 1 mole of $Pb(NO_3)_2$ reacts to produce 2 moles of $NaNO_3$. This ratio is crucial for converting between the amounts of these two substances. If the equation were not balanced, the mole ratios would be incorrect, leading to inaccurate results. Balancing chemical equations is a fundamental skill in chemistry, ensuring that the quantitative relationships in a reaction are correctly represented. This step is not just a formality but a necessity for any stoichiometric calculation. Once we have a balanced equation, we can confidently proceed to use the mole ratios for converting between different substances in the reaction.

Step 2: Calculate the Molar Mass of $NaNO_3$

The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate the molar mass of sodium nitrate ($NaNO_3$), we sum the atomic masses of each element in the compound. The atomic masses can be found on the periodic table:

• Sodium (Na): 22.99 g/mol
• Nitrogen (N): 14.01 g/mol
• Oxygen (O): 16.00 g/mol

Therefore, the molar mass of $NaNO_3$ is:

22.99 rac{g}{mol} (Na) + 14.01 rac{g}{mol} (N) + 3 imes 16.00 rac{g}{mol} (O) = 85.00 rac{g}{mol}

Understanding molar mass is fundamental in stoichiometry as it provides the bridge between mass (grams) and moles. Moles are the SI unit for the amount of a substance and are essential for using the stoichiometric ratios from the balanced chemical equation. In this step, accurately calculating the molar mass of $NaNO_3$ is crucial because it will be used to convert the given mass of $NaNO_3$ (425 grams) into moles. This conversion is necessary to relate the amount of $NaNO_3$ produced to the amount of $Pb(NO_3)_2$ required. A mistake in the molar mass calculation will propagate through the rest of the problem, leading to an incorrect final answer. Thus, attention to detail and the correct use of atomic masses from the periodic table are paramount. The molar mass of a compound is a constant value, making it a reliable conversion factor in stoichiometric calculations. Once the molar mass is determined, it can be used in conjunction with the balanced equation to solve a variety of quantitative problems related to the reaction.

Step 3: Convert Grams of $NaNO_3$ to Moles

To convert grams of $NaNO_3$ to moles, we use the molar mass we calculated in the previous step. We are given 425 grams of $NaNO_3$. Using the molar mass of $NaNO_3$ (85.00 g/mol), we can perform the conversion:

rac{425 ext{ grams } NaNO_3}{1} imes rac{1 ext{ mole } NaNO_3}{85.00 ext{ grams } NaNO_3} = 5.00 ext{ moles } NaNO_3

This conversion is a critical step in stoichiometry because it allows us to work with the mole ratios provided by the balanced chemical equation. Grams are a practical unit for measuring mass in the laboratory, but moles are the unit that directly relates to the number of molecules or ions. The molar mass acts as the conversion factor between these two units. In this calculation, we are essentially determining how many 'units' of $NaNO_3$ we have at the molecular level. This information is then used to determine how many 'units' of $Pb(NO_3)_2$ were required in the reaction. The setup of the conversion is important: we multiply by the molar mass fraction (1 mole / 85.00 grams) so that the grams unit cancels out, leaving us with moles. Ensuring the units cancel correctly is a good way to check the setup of the calculation. Once we have the amount of $NaNO_3$ in moles, we can use the balanced equation to find the corresponding amount of $Pb(NO_3)_2$.

Step 4: Use the Stoichiometric Ratio to Find Moles of $Pb(NO_3)_2$

From the balanced chemical equation:

$Pb(NO_3)_{2(aq)} + 2NaBr_{(aq)} \rightarrow PbBr_{2(s)} + 2NaNO_{3(aq)}$

We can see that 1 mole of $Pb(NO_3)_2$ produces 2 moles of $NaNO_3$. This gives us the stoichiometric ratio:

rac{1 ext{ mole } Pb(NO_3)_2}{2 ext{ moles } NaNO_3}

Using this ratio, we can convert moles of $NaNO_3$ to moles of $Pb(NO_3)_2$:

5.00 ext{ moles } NaNO_3 imes rac{1 ext{ mole } Pb(NO_3)_2}{2 ext{ moles } NaNO_3} = 2.50 ext{ moles } Pb(NO_3)_2

The stoichiometric ratio is the heart of stoichiometric calculations. It's derived directly from the coefficients in the balanced chemical equation and provides the quantitative link between reactants and products. In this case, the ratio of 1 mole $Pb(NO_3)_2$ to 2 moles $NaNO_3$ is crucial because it tells us that for every 2 moles of $NaNO_3$ produced, 1 mole of $Pb(NO_3)_2$ was consumed. Applying this ratio correctly is essential for accurate calculations. We multiply the moles of $NaNO_3$ by the ratio, ensuring that the units of $NaNO_3$ cancel out, leaving us with moles of $Pb(NO_3)_2$. This step bridges the gap between the amount of product we are given (425 grams of $NaNO_3$) and the amount of reactant ($Pb(NO_3)_2$) required. The stoichiometric ratio is not just a number; it represents the fundamental chemical relationship between the substances involved in the reaction. Understanding and applying this ratio correctly is a key skill in stoichiometry. Once we have the moles of $Pb(NO_3)_2$, we can convert it back to grams to answer the original question.

Step 5: Calculate the Molar Mass of $Pb(NO_3)_2$

Similar to step 2, we need to calculate the molar mass of lead(II) nitrate ($Pb(NO_3)_2$). We sum the atomic masses of each element in the compound:

• Lead (Pb): 207.2 g/mol
• Nitrogen (N): 14.01 g/mol
• Oxygen (O): 16.00 g/mol

Therefore, the molar mass of $Pb(NO_3)_2$ is:

207.2 rac{g}{mol} (Pb) + 2 imes igg(14.01 rac{g}{mol} (N) + 3 imes 16.00 rac{g}{mol} (O)igg) = 331.22 rac{g}{mol}

Calculating the molar mass of $Pb(NO_3)_2$ is essential for converting moles back into grams, which is the final step in answering the problem. As with the molar mass of $NaNO_3$, this calculation involves summing the atomic masses of each element in the compound, taking into account the number of atoms of each element as indicated by the chemical formula. The molar mass serves as a conversion factor, linking the macroscopic world of grams, which we can measure in the lab, to the microscopic world of moles, which represent the number of molecules or ions. The accurate determination of molar mass is crucial because any error here will directly affect the final answer. In this case, the molar mass of $Pb(NO_3)_2$ is significantly higher than that of $NaNO_3$, reflecting the heavier atomic mass of lead compared to sodium. This difference in molar mass will influence the amount of $Pb(NO_3)_2$ needed to produce a given amount of $NaNO_3$. Once the molar mass is calculated, it can be used to convert the moles of $Pb(NO_3)_2$ (calculated in the previous step) into grams.

Step 6: Convert Moles of $Pb(NO_3)_2$ to Grams

Now, we convert moles of $Pb(NO_3)_2$ to grams using its molar mass (331.22 g/mol):

2.50 ext{ moles } Pb(NO_3)_2 imes rac{331.22 ext{ grams } Pb(NO_3)_2}{1 ext{ mole } Pb(NO_3)_2} = 828.05 ext{ grams } Pb(NO_3)_2

This final conversion is the culmination of the stoichiometric calculation process. We are taking the amount of $Pb(NO_3)_2$ in moles, which we determined using the balanced chemical equation and the stoichiometric ratio, and converting it back to grams, which is the unit requested in the original problem. The molar mass of $Pb(NO_3)_2$ acts as the conversion factor, allowing us to translate between the number of moles and the mass in grams. Multiplying the moles of $Pb(NO_3)_2$ by its molar mass gives us the mass of $Pb(NO_3)_2$ required to produce 425 grams of $NaNO_3$. The units are crucial here: by setting up the calculation with grams in the numerator and moles in the denominator, we ensure that the moles unit cancels out, leaving us with grams. This final answer provides the solution to the problem, quantifying the amount of lead(II) nitrate needed for the reaction. This conversion highlights the practical application of stoichiometry, allowing us to calculate the mass of reactants needed or products formed in a chemical reaction. With this calculation complete, we have successfully navigated the stoichiometric pathway from a given mass of product to the required mass of reactant.

Final Answer

Therefore, 828.05 grams of lead(II) nitrate ($Pb(NO_3)_2$) will produce 425 grams of sodium nitrate ($NaNO_3$).

Conclusion

In summary, we have successfully calculated the mass of lead(II) nitrate ($Pb(NO_3)_2$) required to produce 425 grams of sodium nitrate ($NaNO_3$) using stoichiometric principles. This calculation involved several key steps, including verifying the balanced chemical equation, calculating molar masses, converting grams to moles, using stoichiometric ratios, and finally, converting moles back to grams. Each step is crucial for the accuracy of the final result, and understanding the underlying concepts is essential for mastering stoichiometry. Stoichiometry is not just a theoretical exercise; it has practical applications in various fields, including chemistry, biology, and engineering. It allows scientists and engineers to predict the amounts of reactants and products in chemical reactions, which is vital for designing experiments, optimizing industrial processes, and ensuring the safety and efficiency of chemical reactions. The ability to perform stoichiometric calculations accurately is a valuable skill that enables one to make informed decisions and solve real-world problems. This problem serves as a clear example of how stoichiometric calculations can be applied to determine the quantitative relationships between substances in a chemical reaction. By following a systematic approach and paying attention to detail, one can confidently tackle stoichiometric problems and gain a deeper understanding of chemical reactions and their quantitative aspects.