Stoichiometry: Calculating Water Mass From Pentane Combustion

Hey guys! Let's dive into a stoichiometry problem where we'll figure out how much water is produced when pentane ($C_5H_{12}$) reacts with oxygen ($O_2$). This is a classic chemistry problem, and we'll break it down step-by-step so it's super easy to understand. We're given that 50 grams of pentane react with 100 grams of oxygen, and we need to find out the mass of water ($H_2O$) produced. Buckle up, and let's get started!

a. Balancing the Chemical Equation

First things first, we need to balance the chemical equation. Balancing equations ensures that we follow the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means the number of atoms for each element must be the same on both sides of the equation. The unbalanced equation we have is:

$C_5H_{12} + O_2 ightarrow CO_2 + H_2O$

Let’s balance this step by step:

1. Balance Carbon (C): There are 5 carbon atoms on the left ($C_5H_{12}$), so we need 5 carbon dioxide ($CO_2$) molecules on the right:

   $C_5H_{12} + O_2 ightarrow 5CO_2 + H_2O$

2. Balance Hydrogen (H): There are 12 hydrogen atoms on the left ($C_5H_{12}$), so we need 6 water ($H_2O$) molecules on the right:

   $C_5H_{12} + O_2 ightarrow 5CO_2 + 6H_2O$

3. Balance Oxygen (O): Now, let's count the oxygen atoms on the right side. We have 5 $CO_2$ molecules, each with 2 oxygen atoms, and 6 $H_2O$ molecules, each with 1 oxygen atom. That’s a total of (5 * 2) + 6 = 16 oxygen atoms. So, we need 8 oxygen molecules ($8O_2$) on the left:

   $C_5H_{12} + 8O_2 ightarrow 5CO_2 + 6H_2O$

So, the balanced equation is:

$C_5H_{12} + 8O_2 ightarrow 5CO_2 + 6H_2O$

The coefficients for the balanced reaction are 1 for $C_5H_{12}$, 8 for $O_2$, 5 for $CO_2$, and 6 for $H_2O$. These coefficients are crucial because they tell us the molar ratios in which the reactants combine and the products are formed. For every 1 mole of pentane that reacts, 8 moles of oxygen are needed, and 5 moles of carbon dioxide and 6 moles of water are produced. Understanding these ratios is the key to solving stoichiometry problems.

b. Calculating Moles of Reactants

Next, we need to find out how many moles of each reactant we have. To do this, we'll use the formula:

Moles = Mass / Molar Mass

1. Moles of Pentane ($C_5H_{12}$)

The molar mass of pentane ($C_5H_{12}$) can be calculated by adding the atomic masses of its constituent atoms:

• Carbon (C): 5 atoms * 12.01 g/mol = 60.05 g/mol
• Hydrogen (H): 12 atoms * 1.01 g/mol = 12.12 g/mol

Molar mass of $C_5H_{12}$ = 60.05 + 12.12 = 72.17 g/mol

Now, we can calculate the moles of pentane:

Moles of $C_5H_{12}$ = 50 g / 72.17 g/mol ≈ 0.693 moles

2. Moles of Oxygen ($O_2$)

The molar mass of oxygen ($O_2$) is:

• Oxygen (O): 2 atoms * 16.00 g/mol = 32.00 g/mol

Now, we calculate the moles of oxygen:

Moles of $O_2$ = 100 g / 32.00 g/mol = 3.125 moles

So, we have approximately 0.693 moles of pentane and 3.125 moles of oxygen. These values are essential for determining the limiting reactant and calculating the mass of the products formed. The limiting reactant is the reactant that is completely consumed in the reaction, and it dictates the maximum amount of product that can be formed. Identifying the limiting reactant is a critical step in stoichiometry.

c. Determining the Limiting Reactant

To figure out the limiting reactant, we need to compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. From the balanced equation, we know that 1 mole of $C_5H_{12}$ reacts with 8 moles of $O_2$.

Let's calculate how much oxygen we would need to react completely with the amount of pentane we have:

Moles of $O_2$ needed = Moles of $C_5H_{12}$ * 8 = 0.693 moles * 8 ≈ 5.544 moles

We only have 3.125 moles of $O_2$, which is less than the 5.544 moles needed to react completely with the pentane. This means that oxygen ($O_2$) is the limiting reactant. The amount of water produced will be determined by the amount of oxygen available, not the pentane.

Since $O_2$ is the limiting reactant, we'll use its moles to calculate the mass of water produced.

Calculating the Mass of Water ($H_2O$) Produced

From the balanced equation, we see that 8 moles of $O_2$ produce 6 moles of $H_2O$. We can set up a proportion to find out how many moles of water are produced from 3.125 moles of $O_2$:

(Moles of $H_2O$ / Moles of $O_2$) = (6 moles $H_2O$ / 8 moles $O_2$)

Moles of $H_2O$ = (3.125 moles $O_2$ * 6 moles $H_2O$) / 8 moles $O_2$ = 2.34375 moles

Now that we know the moles of water produced, we can calculate the mass of water using the formula:

Mass = Moles * Molar Mass

The molar mass of water ($H_2O$) is:

• Hydrogen (H): 2 atoms * 1.01 g/mol = 2.02 g/mol
• Oxygen (O): 1 atom * 16.00 g/mol = 16.00 g/mol

Molar mass of $H_2O$ = 2.02 + 16.00 = 18.02 g/mol

Mass of $H_2O$ = 2.34375 moles * 18.02 g/mol ≈ 42.23 grams

Therefore, approximately 42.23 grams of water are produced in this reaction. This is our final answer! By carefully following the steps of balancing the equation, calculating moles, identifying the limiting reactant, and using stoichiometric ratios, we've successfully determined the mass of water produced.

Summary of Steps

To recap, here are the steps we followed to solve this stoichiometry problem:

1. Balance the chemical equation: We balanced the equation $C_5H_{12} + O_2 ightarrow CO_2 + H_2O$ to get $C_5H_{12} + 8O_2 ightarrow 5CO_2 + 6H_2O$.
2. Calculate moles of reactants: We found that we had approximately 0.693 moles of $C_5H_{12}$ and 3.125 moles of $O_2$.
3. Determine the limiting reactant: We identified $O_2$ as the limiting reactant because it would run out before $C_5H_{12}$.
4. Calculate moles of water produced: Using the stoichiometric ratios, we determined that 3.125 moles of $O_2$ would produce approximately 2.34375 moles of $H_2O$.
5. Calculate mass of water produced: Finally, we calculated that 2.34375 moles of $H_2O$ is equivalent to approximately 42.23 grams of $H_2O$.

Conclusion

So, guys, we've successfully calculated the mass of water produced from the combustion of pentane! Stoichiometry can seem tricky at first, but by breaking it down into manageable steps, it becomes much easier. Remember to always balance your equations, calculate moles, find the limiting reactant, and use those lovely stoichiometric ratios. Keep practicing, and you'll be a stoichiometry pro in no time! If you have any more questions or want to dive deeper into other chemistry topics, just let me know. Happy calculating!