Calculating the Mass of Oxygen Required to React with Methane: A Comprehensive Guide

To determine the mass of oxygen required to react with 70 grams of methane, we must start with a balanced chemical equation. The combustion of methane is a well-known chemical reaction that can be represented as follows:

1. The Balanced Chemical Equation for Methane Combustion

The balanced equation for the combustion of methane (CH4) is:

CH4 2O2 → CO2 2H2O

From this equation, we see that one mole of methane reacts with two moles of oxygen.

2. Calculating the Moles of Methane

To proceed, we need to convert the mass of methane into moles using its molar mass.

Step 1: Molar Mass of Methane (CH4)

Carbon (C): 12.01 g/mol Hydrogen (H): 1.008 g/mol × 4 4.032 g/mol Total molar mass of CH4: 12.01 g/mol 4.032 g/mol 16.042 g/mol

Now, let's calculate the moles of methane in 70 grams:

[ moles frac{mass}{molar mass} frac{70 g}{16.042 g/mol} approx 4.36 mol ]

3. Calculating the Moles of Oxygen Needed

From the balanced equation, we know that 1 mole of CH4 requires 2 moles of O2. Therefore, we can calculate the moles of oxygen needed:

[ moles_{O2} 2 times moles_{CH4} 2 times 4.36 mol approx 8.72 mol ]

4. Calculating the Mass of Oxygen

The molar mass of oxygen (O) is 16.00 g/mol × 2 32.00 g/mol. Now, we can determine the mass of oxygen required:

[ mass_{O2} moles times molar mass 8.72 mol times 32.00 g/mol approx 278.94 g ]

Conclusion: The Mass of Oxygen Required

The mass of oxygen required to react with 70 grams of methane is approximately 279 grams.

Additional Example: Calculating for 32 grams of Methane

Here is another example for calculating the mass of oxygen required, this time for 32 grams of methane:

Step 1: Calculate the Moles of CH4

Using the molar mass of methane (CH4 16 g/mol):

[ moles_{CH4} frac{32 g}{16 g/mol} 2.0 mol ]

Step 2: Calculate the Moles of O2

The mole ratio between CH4 and O2 in the balanced equation is 1:2:

[ moles_{O2} 2.0 mol times 2 4.0 mol ]

Step 3: Calculate the Mass of O2

The molar mass of oxygen (O2) 32 g/mol:

[ mass_{O2} 4.0 mol times 32 g/mol 64 g ]

Conclusion for the Example

The mass of oxygen required to react with 32 grams of methane is 64 grams.

Another Example: Combining the Concepts

For 32 grams of methane, let's re-calculate using the full steps:

[ moles_{CH4} frac{32 g}{16 g/mol} 2.0 mol ]

From the balanced equation, the mole ratio is 1:2, so:

[ moles_{O2} 2 times 2.0 mol 4.0 mol ]

The molar mass of O2 32 g/mol:

[ mass_{O2} 4.0 mol times 32 g/mol 96 g ]

Final Conclusion: Combined Approach

The mass of oxygen required to react with 32 grams of methane is 96 grams.

This comprehensive guide covers the step-by-step process of calculating the required mass of oxygen to react with a given amount of methane, using the principles of balanced chemical equations, molar mass, and mole ratios.