Use the information from the chemical reaction below to determine the amount of sulfur needed to produce 3.52 g of lead sulfide. Hint: Hint: Make sure to identify the conversion factor that can be used to convert g sulfur to g lead sulfide from the given proportions in the reaction. 10.00g of lead + 1.55g of sulfur > 11.55g of lead sulfate.

We are asked to determine the amount of sulfur needed to produce 3.52 g of lead sulfide (PbS). Here's the approach we can use:

### Step 1: Write the chemical equation
From the information provided, we know that:
- 10.00 g of lead (Pb) reacts with 1.55 g of sulfur (S) to produce 11.55 g of lead sulfide (PbS).

This reaction can be represented as:
\[
\text{Pb} + \text{S} \rightarrow \text{PbS}
\]

### Step 2: Understand the ratio of sulfur to lead sulfide
From the provided dataWe are asked to determine the amount of sulfur needed to produce 3.52 g of lead sulfide (PbS). Here's the approach we can use:

### Step 1: Write the chemical equation
From the information provided, we know that:
- 10.00 g of lead (Pb) reacts with 1.55 g of sulfur (S) to produce 11.55 g of lead sulfide (PbS).

This reaction can be represented as:
\[
\text{Pb} + \text{S} \rightarrow \text{PbS}
\]

### Step 2: Understand the ratio of sulfur to lead sulfide
From the provided 
- 1.55 g of sulfur produces 11.55 g of lead sulfide.

This gives us the conversion factor between sulfur and lead sulfide. The mass ratio between sulfur and lead sulfide is:
\[
\frac{1.55 \, \text{g S}}{11.55 \, \text{g PbS}}
\]

### Step 3: Use the conversion factor to find the mass of sulfur required
Now, we want to find how much sulfur is needed to produce 3.52 g of lead sulfide. Using the conversion factor, we can calculate the mass of sulfur:
\[
\text{Mass of S} = 3.52 \, \text{g PbS} \times \frac{1.55 \, \text{g S}}{11.55 \, \text{g PbS}}
\]

Let's compute that.

\[
\text{Mass of S} = 3.52 \times \frac{1.55}{11.55}
\]The amount of sulfur required to produce 3.52 g of lead sulfide is approximately **0.472 g**.

Would you like further details or have any additional questions?

Here are five related questions to expand on this concept:
1. How would you calculate the amount of lead required to produce 3.52 g of lead sulfide?
2. Can this method be used to determine the limiting reagent in a reaction?
3. What is the molar mass of lead sulfide, and how does it relate to stoichiometric calculations?
4. How can you convert the amount of sulfur required from grams to moles?
5. How does the law of conservation of mass apply to this reaction?

**Tip:** Always ensure that your conversion factors are based on accurate mass ratios or molar ratios from balanced chemical equations.

Use the information from the chemical reaction below to determine the amount of sulfur needed to produce 3.52 g of lead sulfide. Hint: Make sure to identify the conversion factor that can be used to convert g sulfur to g lead sulfide from the given proportions in the reaction. 10.00 g of lead + 1.55 g of sulfur > 11.55 g of lead sulfide.

Learn how to determine the mass of sulfur needed to produce 3.52 g of lead sulfide using stoichiometric principles. This step-by-step solution covers the use of mass proportions and the law of conservation of mass. Suitable for high school students in Grades 10-12.

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