Understanding Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. In this problem, we are looking at the combustion of sulfur to form sulfur dioxide ($SO_2$):
$S + O_2 \rightarrow SO_2$
From the balanced equation, we can see that one atom of sulfur reacts with one molecule of oxygen gas to produce one molecule of $SO_2$. Scaling this up to the molar level, 1 mole of sulfur atoms reacts with 1 mole of oxygen molecules to yield 1 mole of $SO_2$ molecules.
Step-by-Step Solution
1. Identify the Molar Relationships
First, we need to find how many moles of $SO_2$ are equivalent to $6.022 \times 10^{24}$ molecules. We know that Avogadro's constant ($N_A$) is $6.022 \times 10^{23}$ molecules/mole.
$\text{Number of moles (n)} = \frac{\text{Total molecules}}{N_A} = \frac{6.022 \times 10^{24}}{6.022 \times 10^{23}} = 10 \text{ moles of } SO_2$
2. Determine Reactants Required
Since the stoichiometric ratio is 1:1:1, producing 10 moles of $SO_2$ requires:
- 10 moles of Sulfur (S)
- 10 moles of Oxygen gas ($O_2$)
3. Convert Moles to Grams
Now, we multiply the number of moles by the molar mass (atomic/molecular weight) of each substance.
For Sulfur (S):
- Atomic mass of S = $32 \text{ g/mol}$
- Mass = $10 \text{ moles} \times 32 \text{ g/mol} = 320 \text{ grams}$
For Oxygen ($O_2$):
- Molar mass of $O_2 = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol}$
- Mass = $10 \text{ moles} \times 32 \text{ g/mol} = 320 \text{ grams}$
Final Result
To produce $6.022 \times 10^{24}$ molecules of $SO_2$, you need 320 grams of sulfur and 320 grams of oxygen gas.