Question

the drawing below shows a mixture of molecules:
key
carbon\thydrogen
nitrogen\tsulfur
oxygen\tchlorine
suppose the following chemical reaction can take place in this mixture:
\\(\ce{cs_{2}(g) + 3 o_{2}(g) \
ightarrow co_{2}(g) + 2 so_{2}(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 each reactant

- For \( \text{CS}_2 \): From the diagram, the molecules with 1 carbon (black) and 2 sulfur (yellow) are \( \text{CS}_2 \). Count them: there are 3 \( \text{CS}_2 \) molecules, so moles of \( \text{CS}_2 = 3 \).
- For \( \text{O}_2 \): The red diatomic molecules are \( \text{O}_2 \). Count them: there are 9 \( \text{O}_2 \) molecules, so moles of \( \text{O}_2 = 9 \).

Step2: Analyze mole ratios from reaction

The reaction is \( \text{CS}_2(g) + 3\text{O}_2(g) ightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \). The mole ratio of \( \text{CS}_2 : \text{O}_2 \) required is \( 1:3 \).

Step3: Determine most, least, and limiting reactant

- Most initial moles: \( \text{O}_2 \) has 9 moles, \( \text{CS}_2 \) has 3 moles. So \( \text{O}_2 \) has the most.
- Least initial moles: \( \text{CS}_2 \) has 3 moles, less than \( \text{O}_2 \)'s 9. So \( \text{CS}_2 \) has the least.
- Limiting reactant: For 3 moles of \( \text{CS}_2 \), we need \( 3 \times 3 = 9 \) moles of \( \text{O}_2 \) (from ratio \( 1:3 \)). We have exactly 9 moles of \( \text{O}_2 \), so both reactants are consumed completely? Wait, no—wait, initial moles: \( \text{CS}_2 = 3 \), \( \text{O}_2 = 9 \). The ratio needed is \( 1:3 \), so \( 3 \text{CS}_2 \) needs \( 9 \text{O}_2 \). So both are in stoichiometric ratio? Wait, maybe I miscounted \( \text{CS}_2 \). Wait the diagram: the yellow (sulfur) and black (carbon) molecules: let's re - count. The \( \text{CS}_2 \) molecules: each \( \text{CS}_2 \) is 1 C and 2 S. Looking at the left box: how many \( \text{CS}_2 \)? Let's see: first row: 1 (the yellow - black - yellow), second row: 1 (yellow - black - yellow), third row: 1 (yellow - black - yellow). Wait, that's 3? Wait no, maybe I made a mistake. Wait the key: red is oxygen (diatomic \( \text{O}_2 \)), yellow is sulfur, black is carbon. So \( \text{CS}_2 \) is a molecule with 1 black (C) and 2 yellow (S), so each \( \text{CS}_2 \) is a tri - atomic molecule (C and two S). Let's count \( \text{CS}_2 \): in the left box, how many? Let's see: first, the molecules: the red ones are \( \text{O}_2 \) (9), and the yellow - black - yellow ones: let's count again. Wait, maybe I miscounted \( \text{CS}_2 \). Wait the original diagram: let's see the left box:

Looking at the left panel:

- Red (O₂): Let's count: first column: 3 (top, middle, bottom), second column: 2 (middle, bottom), third column: 2 (top, middle), fourth column: 2 (middle, bottom)? Wait no, maybe better to count the \( \text{CS}_2 \) (the ones with C and S) and \( \text{O}_2 \) (red diatomic).

Wait the user's diagram: the left box has:

- \( \text{CS}_2 \) molecules: let's see, the ones with black (C) and two yellow (S): I see 3 of them (maybe I was right earlier). And \( \text{O}_2 \) (red diatomic): let's count: 9 (as 9 red pairs).

So moles of \( \text{CS}_2 = 3 \), moles of \( \text{O}_2 = 9 \).

The reaction ratio is \( \text{CS}_2 : \text{O}_2 = 1:3 \). So for 3 moles of \( \text{CS}_2 \), we need \( 3\times3 = 9 \) moles of \( \text{O}_2 \), which is exactly what we have. So both reactants are consumed, but in terms of "limiting reactant", when the mole ratio is exactly the stoichiometric ratio, both can be considered, but wait, maybe I made a mistake in counting \( \text{CS}_2 \). Wait maybe there are 3 \( \text{CS}_2 \) and 9 \( \text{O}_2 \).

So:

- Most initial moles: \( \text{O}_2 \) (9 moles)
- Least initial moles: \( \text{CS}_2 \) (3 moles)
- Limiting reactant: Since the ratio of \( \text{CS}_2 \) to \( \text{O}_2 \) is \( 3:9 = 1:3 \), which is the stoichiometric ratio, both are consumed, but in the reaction, the limiting reactant is the one that runs out first. But since they are in exact ratio, both are limiting? Wait no, maybe I miscounted \( \text{CS}_2 \). Wait maybe there are 3 \( \text{CS}_2 \) and 9 \( \text{O}_2 \). So for each \( \text{CS}_2 \), we need 3 \( \text{O}_2 \). So 3 \( \text{CS}_2 \) need 9 \( \text{O}_2 \), which is available. So both are completely reacted, but the limiting reactant is the one that determines the amount of product. In this case, since they are in stoichiometric ratio, either could be considered, but wait, maybe I made a mistake in counting \( \text{CS}_2 \). Wait maybe the number of \( \text{CS}_2 \) is 3 and \( \text{O}_2 \) is 9. So:

Most initial moles: \( \text{O}_2 \)

Least initial moles: \( \text{CS}_2 \)

Limiting reactant: \( \text{CS}_2 \) (because if we had more \( \text{CS}_2 \), we would need more \( \text{O}_2 \), but here \( \text{O}_2 \) is exactly enough for 3 \( \text{CS}_2 \))

Answer:

Most initial moles: \( \text{O}_2 \)

Least initial moles: \( \text{CS}_2 \)

Limiting reactant: \( \text{CS}_2 \)