Certainly, let's go through the solution step-by-step for each part of the question.
### Step 1: Calculate the number of moles of each reactant
First, we need to find out how many moles of Zinc (Zn) and Sulfur (S) we have initially.
[tex]\[
\text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} = \frac{25\text{ g}}{65\text{ g/mol}} = 0.3846\text{ mol of Zn}
\][/tex]
[tex]\[
\text{Moles of S} = \frac{\text{Mass of S}}{\text{Molar mass of S}} = \frac{30\text{ g}}{32\text{ g/mol}} = 0.9375\text{ mol of S}
\][/tex]
### Step 2: Determine the limiting reactant
The chemical reaction is:
[tex]\[
Zn + S \rightarrow ZnS
\][/tex]
This reaction proceeds in a 1:1 molar ratio between Zn and S. Therefore, the limiting reactant will be the one with the lesser number of moles.
Comparing the moles of Zn (0.3846 mol) and S (0.9375 mol), we see that Zinc (Zn) is the limiting reactant because it has fewer moles available.
### Step 3: Calculate the grams of ZnS formed
Since the limiting reactant is Zinc (Zn), it will determine the amount of the product ZnS formed. Given that the reaction has a 1:1 molar ratio, the moles of ZnS formed will be equal to the moles of Zn.
[tex]\[
\text{Moles of ZnS} = 0.3846\text{ mol}
\][/tex]
Next, we calculate the mass of ZnS. The molar mass of ZnS is obtained by adding the molar masses of Zn and S:
[tex]\[
\text{Molar mass of ZnS} = \text{Molar mass of Zn} + \text{Molar mass of S} = 65\text{ g/mol} + 32\text{ g/mol} = 97\text{ g/mol}
\][/tex]
Now we calculate the mass of ZnS formed:
[tex]\[
\text{Mass of ZnS} = \text{Moles of ZnS} \times \text{Molar mass of ZnS} = 0.3846\text{ mol} \times 97\text{ g/mol} = 37.31\text{ g}
\][/tex]
### Step 4: Calculate the grams of excess reactant left over
Since Zinc is the limiting reactant, Sulfur (S) is the excess reactant. To find out how much Sulfur is left, we first determine how much Sulfur reacts:
[tex]\[
\text{Moles of S reacted} = \text{Moles of the limiting reactant, Zn} = 0.3846\text{ mol}
\][/tex]
We calculate the initial moles of Sulfur (S) and subtract the amount that reacted:
[tex]\[
\text{Moles of excess S} = \text{Initial moles of S} - \text{Moles of S reacted} = 0.9375\text{ mol} - 0.3846\text{ mol} = 0.5529\text{ mol}
\][/tex]
Now, we convert these remaining moles of Sulfur back to grams using its molar mass:
[tex]\[
\text{Mass of excess S} = \text{Moles of excess S} \times \text{Molar mass of S} = 0.5529\text{ mol} \times 32\text{ g/mol} = 17.69\text{ g}
\][/tex]
### Summary of Answers:
a) The limiting reactant is Zinc (Zn).
b) The grams of ZnS formed is 37.31 g.
c) The grams of excess Sulfur (S) left over is 17.69 g.
