Question

the drawing below shows a mixture of molecules. suppose the following chemical reaction can take place in this mixture: 2c₂h₂(g)+5o₂(g)→4co₂(g)+2h₂o(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 reactant molecules

Count the number of $C_2H_2$ and $O_2$ molecules in the mixture. Assume black - carbon, red - oxygen, and white - hydrogen. By counting, we find the number of $C_2H_2$ and $O_2$ molecules.

Step2: Determine the mole - ratio from the balanced equation

The balanced chemical equation is $2C_2H_2(g)+5O_2(g)
ightarrow4CO_2(g) + 2H_2O(g)$. The mole - ratio of $C_2H_2$ to $O_2$ is $\frac{n_{C_2H_2}}{n_{O_2}}=\frac{2}{5}=0.4$.

Step3: Calculate the mole - ratio of the reactants in the mixture

Let's say we count $x$ number of $C_2H_2$ molecules and $y$ number of $O_2$ molecules in the mixture. Calculate the ratio $\frac{x}{y}$. If $\frac{x}{y}<0.4$, $C_2H_2$ is the limiting reactant; if $\frac{x}{y}>0.4$, $O_2$ is the limiting reactant.

Step4: Identify the reactant with the most and least moles

After counting the molecules of each reactant, we can directly identify which reactant has the most and least initial moles.

Let's assume we count 4 molecules of $C_2H_2$ and 12 molecules of $O_2$.
The mole - ratio of the reactants in the mixture is $\frac{4}{12}=\frac{1}{3}\approx0.33<0.4$. So $C_2H_2$ is the limiting reactant.
The number of $O_2$ molecules (12) is more than the number of $C_2H_2$ molecules (4).

Answer:

Limiting reactant: $C_2H_2$
Reactant with the least initial moles: $C_2H_2$
Reactant with the most initial moles: $O_2$