Calculating Grams Of P₄O₁₀ Needed To React With 9 Moles Of H₂O

Introduction to P₄O₁₀ and Its Reaction with Water

In the realm of chemistry, understanding stoichiometry is crucial for predicting the outcomes of chemical reactions. Stoichiometry deals with the quantitative relationships between reactants and products in chemical reactions. A classic example of a stoichiometric calculation involves the reaction between phosphorus pentoxide ($P_4O_{10}$) and water ($H_2O$). This reaction is highly exothermic and results in the formation of phosphoric acid ($H_3PO_4$). Let's delve into the specifics of this reaction and explore how to calculate the amount of $P_4O_{10}$ needed to react with a given amount of $H_2O$.

The chemical equation for the reaction between phosphorus pentoxide and water is:

$P_4 O_{10} + 6 H_2 O ightarrow 4 H_3 PO_4$

This equation tells us that one mole of $P_4O_{10}$ reacts with six moles of $H_2O$ to produce four moles of $H_3PO_4$. This mole ratio is the key to solving stoichiometric problems related to this reaction. To accurately determine the mass of $P_4O_{10}$ required, we must first understand the molar masses of the compounds involved. The molar mass of a compound is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). In this article, we will embark on a detailed journey to calculate the exact mass of $P_4O_{10}$ required to react completely with 9 moles of $H_2O$, providing a step-by-step guide suitable for students and chemistry enthusiasts alike. By mastering such calculations, you’ll gain a deeper insight into the quantitative nature of chemical reactions, making it easier to predict and control chemical processes in both laboratory and industrial settings. We will emphasize the importance of understanding mole ratios, molar masses, and the balanced chemical equation to ensure accurate stoichiometric calculations. So, let’s proceed with the molar mass calculations as the first step in unraveling this chemical puzzle.

Calculating Molar Masses

To tackle the question of how many grams of $P_4O_{10}$ are needed, we first need to calculate the molar masses of the reactants involved: $P_4O_{10}$ and $H_2O$. The molar mass of a compound is the sum of the atomic masses of all the atoms in its formula. We are given the atomic masses of phosphorus (P), oxygen (O), and hydrogen (H) as follows:

• P = 30.97 g/mol
• O = 15.999 g/mol
• H = 1.008 g/mol

Let’s start by calculating the molar mass of phosphorus pentoxide ($P_4O_{10}$). The molecule consists of 4 phosphorus atoms and 10 oxygen atoms. Therefore, the molar mass of $P_4O_{10}$ can be calculated as:

Molar mass of $P_4O_{10}$ = (4 × Atomic mass of P) + (10 × Atomic mass of O) Molar mass of $P_4O_{10}$ = (4 × 30.97 g/mol) + (10 × 15.999 g/mol) Molar mass of $P_4O_{10}$ = 123.88 g/mol + 159.99 g/mol Molar mass of $P_4O_{10}$ = 283.87 g/mol

Next, we need to calculate the molar mass of water ($H_2O$). A water molecule consists of 2 hydrogen atoms and 1 oxygen atom. Thus, the molar mass of $H_2O$ can be calculated as:

Molar mass of $H_2O$ = (2 × Atomic mass of H) + (1 × Atomic mass of O) Molar mass of $H_2O$ = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) Molar mass of $H_2O$ = 2.016 g/mol + 15.999 g/mol Molar mass of $H_2O$ = 18.015 g/mol

Now that we have the molar masses of both $P_4O_{10}$ and $H_2O$, we are well-equipped to proceed with the stoichiometric calculations. Understanding these molar masses is a fundamental step in determining the mass relationships in chemical reactions. Without accurate molar masses, the subsequent calculations would be flawed, leading to incorrect results. So, with these values in hand, we can now delve into using the balanced chemical equation to find out how many moles of $P_4O_{10}$ are required to react with 9 moles of $H_2O$. This step will bring us closer to our final answer: the mass of $P_4O_{10}$ needed.

Using the Balanced Chemical Equation

The balanced chemical equation provides the stoichiometric coefficients, which are the key to understanding the mole ratios between reactants and products. In our case, the balanced equation is:

$P_4 O_{10} + 6 H_2 O ightarrow 4 H_3 PO_4$

This equation tells us that 1 mole of $P_4O_{10}$ reacts with 6 moles of $H_2O$. This is our mole ratio. We are given that we have 9 moles of $H_2O$, and we want to find out how many moles of $P_4O_{10}$ are needed to react with this amount of water. To do this, we use the mole ratio from the balanced equation:

(Moles of $P_4O_{10}$) / (Moles of $H_2O$) = 1 / 6

We can set up a proportion to solve for the moles of $P_4O_{10}$:

Moles of $P_4O_{10}$ / 9 moles $H_2O$ = 1 / 6

Now, we solve for the moles of $P_4O_{10}$:

Moles of $P_4O_{10}$ = (1 / 6) × 9 moles $H_2O$ Moles of $P_4O_{10}$ = 1.5 moles

So, we need 1.5 moles of $P_4O_{10}$ to react completely with 9 moles of $H_2O$. This step is crucial because it bridges the gap between the amount of water we have and the amount of phosphorus pentoxide we need. The stoichiometric coefficients in the balanced equation act as a conversion factor, allowing us to move from moles of one substance to moles of another. Without this balanced equation, we would not be able to accurately determine the required amount of $P_4O_{10}$. Now that we know the number of moles of $P_4O_{10}$ needed, we can move on to the final step: converting moles to grams. This conversion will give us the answer in the units the question asks for, completing our stoichiometric calculation. Understanding the mole concept and how it relates to balanced chemical equations is paramount in chemistry, enabling us to predict and quantify chemical reactions accurately. This careful, step-by-step approach ensures that our calculations are precise and reliable.

Converting Moles to Grams

Now that we know we need 1.5 moles of $P_4O_{10}$, the final step is to convert this amount from moles to grams. To do this, we use the molar mass of $P_4O_{10}$, which we calculated earlier as 283.87 g/mol. The conversion formula is:

Mass = Moles × Molar mass

Plugging in the values we have:

Mass of $P_4O_{10}$ = 1.5 moles × 283.87 g/mol Mass of $P_4O_{10}$ = 425.805 g

Therefore, 425.805 grams of $P_4O_{10}$ are needed to react with 9 moles of $H_2O$. This calculation completes our stoichiometric problem, providing the answer in the desired unit: grams. Converting moles to grams is a fundamental skill in chemistry, allowing us to translate between the mole concept—which is essential for understanding chemical reactions at the molecular level—and the practical masses we measure in the laboratory. The molar mass acts as the conversion factor between these two units, making it a crucial value to have. In this case, by multiplying the number of moles of $P_4O_{10}$ by its molar mass, we were able to determine the exact mass required for the reaction. This process highlights the importance of a systematic approach to solving stoichiometric problems: first, ensuring the chemical equation is balanced; second, using the mole ratio from the balanced equation to find the number of moles of the required reactant; and third, converting moles to grams using the molar mass. With this final calculation, we have successfully answered the question, providing a clear and accurate result. Understanding these steps is key to mastering stoichiometry and confidently tackling similar problems in chemistry.

Conclusion

In summary, to determine how many grams of $P_4O_{10}$ are needed to react with 9 moles of $H_2O$, we followed a systematic stoichiometric approach. First, we established the balanced chemical equation for the reaction:

$P_4 O_{10} + 6 H_2 O ightarrow 4 H_3 PO_4$

Next, we calculated the molar masses of $P_4O_{10}$ and $H_2O$ using the given atomic masses of phosphorus, oxygen, and hydrogen. This step was crucial for converting between moles and grams. We found the molar mass of $P_4O_{10}$ to be 283.87 g/mol and the molar mass of $H_2O$ to be 18.015 g/mol.

Using the balanced equation, we determined the mole ratio between $P_4O_{10}$ and $H_2O$ to be 1:6. This ratio allowed us to calculate that 1.5 moles of $P_4O_{10}$ are required to react with 9 moles of $H_2O$.

Finally, we converted the moles of $P_4O_{10}$ to grams using its molar mass: Mass = Moles × Molar mass. This calculation gave us the final answer: 425.805 grams of $P_4O_{10}$ are needed to react with 9 moles of $H_2O$.

This exercise underscores the importance of stoichiometry in chemistry. Stoichiometry allows us to quantitatively understand chemical reactions, predict the amounts of reactants and products involved, and perform accurate calculations in both laboratory and industrial settings. Mastering stoichiometric principles, such as balancing chemical equations, calculating molar masses, using mole ratios, and converting between moles and grams, is fundamental to success in chemistry. By breaking down complex problems into manageable steps, we can confidently tackle a wide range of chemical calculations. Stoichiometry not only provides numerical answers but also enhances our understanding of the fundamental laws governing chemical reactions. The ability to accurately perform stoichiometric calculations is a cornerstone of chemical knowledge, empowering us to design experiments, optimize chemical processes, and deepen our appreciation of the molecular world. This comprehensive approach ensures that we grasp not just the mechanics of the calculation, but also the underlying chemical principles that make it possible.