Q.How many moles of methane are required to produce 22 g $CO_2$

The key idea is Stoichiometry and Molar Mass Calculation.

1. First, write the balanced chemical equation for the combustion of methane:

This equation shows that 1 mole of $CH_4$ produces 1 mole of $CO_2$.

2. Next, calculate the molar mass of $CO_2$:

For practical calculations, we often use $44 \text{ g/mol}$.

3. Now, determine the number of moles of $CO_2$ produced from the given mass: …

When we talk about chemical reactions, the amounts of reactants consumed and products formed are not arbitrary. They are governed by the fundamental principle of conservation of mass, which means atoms are neither created nor destroyed in a chemical reaction. This principle is mathematically represented by a balanced chemical equation.

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It allows us to predict how much of one substance is needed or produced given the amount of another substance involved in the reaction. The key to stoichiometry is understanding the mole concept and how it connects mass to the number of particles, and how the coefficients in a balanced equation relate the moles of different substances.

In this problem, we are given the mass of a product ($CO_2$) and asked to find the moles of a reactant ($CH_4$) required to produce it. The general strategy involves these steps:

1. Write and balance the chemical equation for the reaction.
2. Convert the given mass of the known substance into moles using its molar mass.
3. Use the mole ratio from the balanced equation to convert moles of the known substance to moles of the unknown substance.

Let's apply this step-by-step.

1. Write and balance the chemical equation for the combustion of methane.

   Combustion is a reaction with oxygen, typically producing carbon dioxide and water. Methane ($CH_4$) is a hydrocarbon.

   The unbalanced equation is:

   $CH_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)$

   To balance it, we ensure the number of atoms of each element is the same on both sides of the equation.

   • Carbon: 1 on left, 1 on right. (Balanced)
   • Hydrogen: 4 on left, 2 on right. Place a coefficient of 2 in front of $H_2O$ on the right: $CH_4(g) + O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$
   • Oxygen: 2 on left, $2$ (from $CO_2$) $+ 2 \times 1$ (from $2H_2O$) $= 4$ on right. Place a coefficient of 2 in front of $O_2$ on the left: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$

   The balanced chemical equation is:

This equation tells us that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

2. Identify the known and unknown quantities.

* Known: Mass of $CO_2 = 22 \text{ g}$.

* Unknown: Moles of $CH_4$.

3. Convert the given mass of $CO_2$ to moles.

   To do this, we need the molar mass of $CO_2$.

   • Molar mass of Carbon (C) $\approx 12.01 \text{ g/mol}$
   • Molar mass of Oxygen (O) $\approx 16.00 \text{ g/mol}$

   Molar mass of $CO_2 = (\text{Molar mass of C}) + 2 \times (\text{Molar mass of O})$

   Molar mass of $CO_2 = 12.01 \text{ g/mol} + 2 \times 16.00 \text{ g/mol}$

   Molar mass of $CO_2 = 12.01 \text{ g/mol} + 32.00 \text{ g/mol}$

   Molar mass of $CO_2 = 44.01 \text{ g/mol}$

   Now, convert the mass of $CO_2$ to moles: …

Method: Mole-to-Mole Stoichiometry via Balanced Chemical Equation

This method uses the balanced chemical equation to relate moles of reactant to moles of product.

Step 1: Write and balance the combustion reaction

Methane ($CH_4$) burns in oxygen to produce carbon dioxide and water:

The balanced equation shows:

• 1 mole of $CH_4$ produces 1 mole of $CO_2$

Step 2: Convert given mass of $CO_2$ to moles

Molar mass of $CO_2$:

• C: $12 \, g/mol$
• O: $16 \times 2 = 32 \, g/mol$
• Total: $12 + 32 = 44 \, g/mol$

Step 3: Use mole ratio from the balanced equation

From the equation: $1 \, mol \, CH_4 : 1 \, mol \, CO_2$ …

✓ First, the correct approach

Combustion of methane:

• Molar mass of $CO_2$ = $12 + 2 \times 16 = 44 \text{ g/mol}$
• Moles of $CO_2$ in 22 g = $\frac{22}{44} = 0.5 \text{ mol}$
• From the balanced equation: 1 mol $CH_4$ → 1 mol $CO_2$
• So, moles of $CH_4$ required = 0.5 mol

✗ Mistake 1: Using wrong molar mass of $CO_2$

What students do:

Take molar mass of $CO_2$ as $12 + 16 = 28$ (confusing it with $CO$) or $12 + 16 \times 2 = 44$ but then miscalculate $16 \times 2$ as 32 (correct) but then add 12 incorrectly.

How to avoid:

• Always write the atomic masses clearly: $C = 12$, $O = 16$
• Then compute: $12 + (16 \times 2) = 12 + 32 = 44$
• Double-check by writing the formula with subscripts: $CO_2$ means 1 C + 2 O.

✗ Mistake 2: Forgetting to balance the equation

What students do:

Use the unbalanced equation $CH_4 + O_2 \rightarrow CO_2 + H_2O$ and assume 1:1 mole ratio for everything.

How to avoid:

• Always balance the chemical equation first.
• For methane combustion, the balanced equation is:

• The key ratio: 1 mol $CH_4$ → 1 mol $CO_2$ (so no extra factor needed here, but you must confirm it).

✗ Mistake 3: Using the wrong mole ratio

What students do:

Think 1 mol $CH_4$ gives 2 mol $CO_2$ (confusing with $H_2O$) or use the $O_2$ coefficient.

How to avoid:

• Read the balanced equation coefficient by coefficient.
• Only the substance you are given and the substance you want matter.
• Here: $CH_4$ and $CO_2$ both have coefficient 1 → ratio is 1:1.

✗ Mistake 4: Calculating moles of $CO_2$ incorrectly

What students do:

Divide mass by molar mass in the wrong order: $44 / 22 = 2$ mol instead of $22 / 44 = 0.5$ mol.

How to avoid:

• Use the formula: $\text{moles} = \frac{\text{given mass (g)}}{\text{molar mass (g/mol)}}$ …