If 50.0 grams of zinc are reacted with 300. mL of 3.00 M HCl(aq) and 250. mg of
hydrogen gas are actually produced, what is the %-yield?

To calculate the percent yield of the reaction, we can follow these steps:

1. **Determine the limiting reagent**: Zinc (Zn) or Hydrochloric acid (HCl).
2. **Find the theoretical yield** of hydrogen gas (H₂) based on the limiting reagent.
3. **Calculate the percent yield** using the formula:

\[
\% \text{yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100
\]

### Step 1: Write the balanced chemical equation
The reaction between zinc and hydrochloric acid is:

\[
\text{Zn(s)} + 2 \text{HCl(aq)} \rightarrow \text{ZnCl}_2(\text{aq}) + \text{H}_2(\text{g})
\]

### Step 2: Calculate moles of each reactant
#### Moles of Zinc:
- Molar mass of Zn = 65.38 g/mol
- Mass of Zn = 50.0 g

\[
\text{moles of Zn} = \frac{50.0 \, \text{g}}{65.38 \, \text{g/mol}} = 0.7648 \, \text{mol}
\]

#### Moles of HCl:
- Molarity of HCl = 3.00 M
- Volume of HCl = 300. mL = 0.300 L

\[
\text{moles of HCl} = \text{Molarity} \times \text{Volume} = 3.00 \, \text{M} \times 0.300 \, \text{L} = 0.900 \, \text{mol}
\]

### Step 3: Determine the limiting reagent
From the balanced equation, we see that 1 mole of Zn reacts with 2 moles of HCl. Let’s check how many moles of HCl are required for 0.7648 moles of Zn:

\[
\text{moles of HCl needed} = 0.7648 \, \text{mol Zn} \times 2 = 1.5296 \, \text{mol HCl}
\]

Since we only have 0.900 moles of HCl, **HCl is the limiting reagent**.

### Step 4: Theoretical yield of H₂ gas
From the balanced equation, 2 moles of HCl produce 1 mole of H₂. Since HCl is limiting, let's calculate the theoretical moles of H₂ that can be produced:

\[
\text{moles of H}_2 = \frac{0.900 \, \text{mol HCl}}{2} = 0.450 \, \text{mol H}_2
\]

Now, convert moles of H₂ to grams. The molar mass of H₂ is 2.02 g/mol:

\[
\text{mass of H}_2 (\text{theoretical}) = 0.450 \, \text{mol} \times 2.02 \, \text{g/mol} = 0.909 \, \text{g}
\]

### Step 5: Calculate percent yield
The actual yield is given as 250 mg (0.250 g). Using the percent yield formula:

\[
\% \text{yield} = \left( \frac{0.250 \, \text{g}}{0.909 \, \text{g}} \right) \times 100 = 27.5\%
\]

### Final Answer:
The percent yield of the reaction is **27.5%**.

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Would you like more details on any step? Here are 5 related questions you might find interesting:

1. How do you identify the limiting reagent in a chemical reaction?
2. Why is it important to calculate the percent yield in chemical reactions?
3. What factors can affect the actual yield in a reaction?
4. How do different units (mg, g, L) play a role in yield calculations?
5. What is the significance of molarity in reactions involving solutions?

**Tip:** Always ensure your units match when calculating yield or reacting quantities to avoid errors.

If 50.0 grams of zinc are reacted with 300. mL of 3.00 M HCl(aq) and 250. mg of hydrogen gas are actually produced, what is the %-yield?

Learn how to calculate the percent yield when reacting 50.0 grams of zinc with 300 mL of 3.00 M HCl. Explore key concepts such as stoichiometry, limiting reagents, and theoretical yield with a detailed step-by-step guide. Suitable for high school chemistry students.

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