Question

the drawing below shows a mixture of molecules: suppose the following chemical reaction can take place in this mixture: ch₄(g)+4s(s)→cs₂(g)+2h₂s(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 the number of each reactant molecules

Count the number of $CH_4$ and $S$ molecules in the mixture. Assume we can count them from the drawing. Let's say we count $n_{CH_4}$ number of methane molecules and $n_S$ number of sulfur - atoms (since sulfur is in atomic form here).

Step2: Determine the initial moles relationship

The number of moles is proportional to the number of molecules ($n=\frac{N}{N_A}$, where $n$ is the number of moles, $N$ is the number of molecules and $N_A$ is Avogadro's number). So, the reactant with more molecules has more initial moles.

Step3: Identify the limiting reactant

From the balanced chemical equation $CH_4(g)+4S(s)
ightarrow CS_2(g) + 2H_2S(g)$, the mole - ratio of $CH_4$ to $S$ is $1:4$. Calculate the amount of product that can be formed from each reactant. The reactant that produces the least amount of product is the limiting reactant. Let the number of moles of $CH_4$ be $n_{CH_4}$ and of $S$ be $n_S$. The amount of $CS_2$ produced from $CH_4$ is $n_{CS_2,CH_4}=n_{CH_4}$ and from $S$ is $n_{CS_2,S}=\frac{n_S}{4}$. Compare these two values to find the limiting reactant.

Answer:

(1) Without seeing the actual drawing, we can't give a definite answer. But if we assume we counted and found more $S$ molecules, the answer for "Of which reactant are there the most initial moles?" is $S$.
(2) If we found fewer $CH_4$ molecules, the answer for "Of which reactant are there the least initial moles?" is $CH_4$.
(3) To find the limiting reactant, we calculate the amount of product formed from each reactant as described above. If $n_{CH_4}>\frac{n_S}{4}$, then $S$ is the limiting reactant. If $n_{CH_4}<\frac{n_S}{4}$, then $CH_4$ is the limiting reactant. Without actual values from the drawing, we can't give a definite chemical - formula for the limiting reactant.