2CO + O2 −-> 2CO2

How many liters of carbon monoxide at 25.0oC and 8.56 mmHg are needed to react with 12.30 L of oxygen gas at STP?

To solve this problem, we need to apply the stoichiometry of the reaction and use the Ideal Gas Law to convert between different gas conditions.

### Given Reaction:
\[
2CO + O_2 \rightarrow 2CO_2
\]

### Step 1: Mole ratio
From the balanced equation, we see that **2 moles of CO react with 1 mole of \( O_2 \)**. This gives us a ratio of 2:1 for \( CO \) to \( O_2 \).

### Step 2: Calculate moles of \( O_2 \) at STP
We are given that there are 12.30 liters of \( O_2 \) at standard temperature and pressure (STP). At STP, 1 mole of gas occupies 22.4 liters. Therefore, the number of moles of \( O_2 \) is:

\[
\text{Moles of } O_2 = \frac{12.30 \, L}{22.4 \, L/mol} = 0.5491 \, \text{moles of } O_2
\]

### Step 3: Use the mole ratio to find moles of \( CO \)
Since the ratio of \( CO \) to \( O_2 \) is 2:1, the moles of \( CO \) required are:

\[
\text{Moles of } CO = 2 \times 0.5491 \, \text{moles} = 1.0982 \, \text{moles of } CO
\]

### Step 4: Convert moles of \( CO \) to liters under the given conditions
We now need to find how many liters of \( CO \) are required at 25.0°C and 8.56 mmHg. To do this, we will use the Ideal Gas Law:

\[
PV = nRT
\]

Where:
- \( P \) is the pressure in atm
- \( V \) is the volume in liters
- \( n \) is the number of moles
- \( R \) is the gas constant \( (0.0821 \, L \, atm / mol \, K) \)
- \( T \) is the temperature in Kelvin \( (T = 25.0°C + 273 = 298.0 \, K) \)

First, convert the pressure from mmHg to atm:

\[
P = \frac{8.56 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.01126 \, \text{atm}
\]

Now, solve for \( V \):

\[
V = \frac{nRT}{P} = \frac{1.0982 \, \text{mol} \times 0.0821 \, L \, atm / mol \, K \times 298.0 \, K}{0.01126 \, atm}
\]

Let me calculate the volume.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, you can use the formula below to finish the calculation yourself:

\[
V = \frac{nRT}{P} = \frac{1.0982 \times 0.0821 \times 298.0}{0.01126}
\]

This will give you the volume of carbon monoxide required in liters.

Would you like more details on any step of the process? Here are some related questions to expand on this:

1. How do I convert between different gas units (e.g., mmHg to atm)?
2. What is the Ideal Gas Law, and when do you use it?
3. How do you calculate moles of a gas from its volume?
4. What is STP, and why is it important in gas law calculations?
5. How does temperature and pressure affect gas volume?

**Tip:** When using the Ideal Gas Law, always check your units to ensure pressure is in atm, volume in liters, and temperature in Kelvin.

How many liters of carbon monoxide at 25.0oC and 8.56 mmHg are needed to react with 12.30 L of oxygen gas at STP?

Learn how to calculate the volume of carbon monoxide needed to react with 12.30 L of oxygen at STP, using stoichiometry and the Ideal Gas Law. Understand key concepts like mole ratio, gas laws, and unit conversions. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation