Question

the drawing below shows a mixture of molecules:
key
carbon hydrogen
nitrogen sulfur
oxygen chlorine
suppose the following chemical reaction can take place in this mixture:
$2\\,\mathrm{c}(s) + \mathrm{o}_2(g) \
ightarrow 2\\,\mathrm{co}(g)$
of which reactant are there the most initial moles? enter its chemical formula:
of which reactant are there the least initial moles? enter its chemical formula:
which reactant is the limiting reactant? enter its chemical formula:

Explanation:

Step1: Count moles of reactants

From the diagram:

• Moles of \( \text{C}(s) \): Count the black (carbon) atoms. There are 8 C atoms, so moles of \( \text{C} = \frac{8}{1} = 8 \) (since each C is 1 mole in the reaction context, as \( \text{C} \) is solid, 1 atom = 1 mole here).
• Moles of \( \text{O}_2(g) \): Count the red (oxygen) molecules. Each \( \text{O}_2 \) is 2 red atoms. There are 5 \( \text{O}_2 \) molecules, so moles of \( \text{O}_2 = 5 \).

Step2: Compare moles for "most"

Compare moles of \( \text{C} \) (8) and \( \text{O}_2 \) (5). \( 8 > 5 \), so most moles: \( \text{C} \).

Step3: Compare moles for "least"

\( 5 < 8 \), so least moles: \( \text{O}_2 \).

Step4: Determine limiting reactant

Reaction: \( 2\text{C}(s) + \text{O}_2(g)
ightarrow 2\text{CO}(g) \).
Mole ratio of \( \text{C}:\text{O}_2 = 2:1 \).
For \( \text{O}_2 = 5 \) moles, required \( \text{C} = 2 \times 5 = 10 \) moles. But we have only 8 moles of \( \text{C} \). Wait, no—wait, initial moles: \( \text{C} = 8 \), \( \text{O}_2 = 5 \).
Using \( \text{C} \): Moles of \( \text{O}_2 \) needed = \( \frac{8}{2} = 4 \) moles. We have 5 moles of \( \text{O}_2 \), which is more than 4.
Using \( \text{O}_2 \): Moles of \( \text{C} \) needed = \( 2 \times 5 = 10 \) moles. We have 8 moles of \( \text{C} \), which is less than 10. So \( \text{C} \) is limiting? Wait, no—wait, let's recheck.
Wait, reaction is \( 2\text{C} + \text{O}_2
ightarrow 2\text{CO} \). So 2 moles C react with 1 mole \( \text{O}_2 \).
Moles of \( \text{C} = 8 \), moles of \( \text{O}_2 = 5 \).
Moles of \( \text{C} \) that can react with \( \text{O}_2 \): \( 2 \times 5 = 10 \) (but we have 8, so \( \text{C} \) is limiting? Wait, no—if \( \text{O}_2 = 5 \) moles, it can react with 10 moles of C, but we have only 8. So C is limiting? Wait, no, wait: when \( \text{C} = 8 \) moles, it can react with \( \frac{8}{2} = 4 \) moles of \( \text{O}_2 \). Since we have 5 moles of \( \text{O}_2 \), \( \text{O}_2 \) is in excess, and \( \text{C} \) is limiting? Wait, no, I think I messed up. Wait, the reaction is 2 C + 1 O₂ → 2 CO. So for each mole of O₂, 2 moles of C are needed.
If we have 5 moles of O₂, we need 10 moles of C. But we have only 8 moles of C. So C is the limiting reactant? Wait, no—wait, initial moles: C is 8, O₂ is 5.
So C can react with \( \frac{8}{2} = 4 \) moles of O₂. O₂ available is 5, which is more than 4. So O₂ is in excess, and C is limiting? Wait, no, the limiting reactant is the one that runs out first. So if we use C first: 8 moles C can react with 4 moles O₂ (since 2 C per 1 O₂). So after reaction, C is used up (8 moles), O₂ used 4 moles, remaining O₂ = 5 - 4 = 1 mole. So C is limiting. Wait, but earlier I thought O₂ was least, but let's confirm counts again.

Wait, the diagram: red molecules (O₂) are 5 (each O₂ is two red atoms, so 5 molecules). Black atoms (C) are 8 (each is a C atom, so 8 moles of C). So:

• Most initial moles: \( \text{C} \) (8 moles)
• Least initial moles: \( \text{O}_2 \) (5 moles)
• Limiting reactant: \( \text{C} \)? Wait, no—wait, reaction is 2 C + O₂ → 2 CO. So mole ratio C:O₂ = 2:1.

Moles of C = 8, moles of O₂ = 5.

For C to react completely: moles of O₂ needed = 8 / 2 = 4. We have 5 O₂, so O₂ is excess. So C is limiting.

For O₂ to react completely: moles of C needed = 5 * 2 = 10. We have 8 C, so C is limiting. Yes, so limiting reactant is C.

Answer:

• Most initial moles: \( \boldsymbol{\text{C}} \)
• Least initial moles: \( \boldsymbol{\text{O}_2} \)
• Limiting reactant: \( \boldsymbol{\text{C}} \)

(Note: If the table requires three separate answers, fill each cell with \( \text{C} \), \( \text{O}_2 \), \( \text{C} \) respectively.)