https://file.aiquickdraw.com/mathbot/file/1726711080481afifgliw.jpg

The exercise provided in the image contains stoichiometric problems related to limiting reactants and reactions. Let's break down and solve the first problem step by step:

### Problem 1:

**Given:**
- 10.8 grams of aluminum (Al) react with 49 grams of sulfuric acid (H₂SO₄).
- The reaction is:

\[
2Al(s) + 3H₂SO₄(aq) \rightarrow Al₂(SO₄)₃(aq) + 3H₂(g)
\]

We need to:
1. **Determine the limiting reactant.**
2. **Calculate the moles of the remaining reactant.**
3. **Calculate the volume of hydrogen gas (H₂) produced at STP.**

---

### **Step 1: Molar masses**
- Molar mass of **Al**: \( 26.98 \, \text{g/mol} \)
- Molar mass of **H₂SO₄**: \( 98.08 \, \text{g/mol} \)

### **Step 2: Convert given masses to moles**
- Moles of **Al**:
\[
\frac{10.8 \, \text{g}}{26.98 \, \text{g/mol}} = 0.400 \, \text{mol}
\]
- Moles of **H₂SO₄**:
\[
\frac{49 \, \text{g}}{98.08 \, \text{g/mol}} = 0.499 \, \text{mol}
\]

### **Step 3: Reaction ratio**
The balanced reaction shows the ratio:
\[
2 \, \text{mol Al} : 3 \, \text{mol H₂SO₄}
\]
For 0.400 mol of Al, the required moles of H₂SO₄ are:
\[
\frac{3}{2} \times 0.400 = 0.600 \, \text{mol H₂SO₄}
\]
But we only have 0.499 mol of H₂SO₄, so **H₂SO₄** is the limiting reactant.

---

### **Step 4: Calculate the moles of remaining reactant (Al)**
Since H₂SO₄ is the limiting reactant, we calculate the amount of Al that reacts:
\[
\text{Required moles of Al} = \frac{2}{3} \times 0.499 = 0.333 \, \text{mol Al}
\]
The remaining moles of Al:
\[
0.400 - 0.333 = 0.067 \, \text{mol Al}
\]

---

### **Step 5: Calculate the volume of H₂ gas at STP**
From the balanced equation, we know 3 moles of H₂ are produced for every 2 moles of Al. Thus, for 0.499 mol of H₂SO₄, the moles of H₂ produced:
\[
\frac{3}{3} \times 0.499 = 0.499 \, \text{mol H₂}
\]
At STP, 1 mol of any gas occupies 22.4 liters. Therefore, the volume of H₂ gas produced:
\[
0.499 \, \text{mol} \times 22.4 \, \text{L/mol} = 11.18 \, \text{L}
\]

---

### Final answers for Problem 1:
a. The limiting reactant is **H₂SO₄**.  
b. The remaining moles of Al are **0.067 mol**.  
c. The volume of H₂ gas produced at STP is **11.18 L**.

---

Would you like a detailed explanation of the second problem as well? Here are 5 related questions to explore:

1. How do you calculate the molar mass of a compound?
2. What is the significance of a limiting reactant in a chemical reaction?
3. How does the gas volume change at non-STP conditions?
4. What are the common methods to identify the limiting reactant?
5. How do you convert between grams and moles in chemical reactions?

**Tip**: Always balance your chemical equation first before solving stoichiometric problems. It ensures the correct mole ratios are used!

Sebanyak 10,8 gram logam aluminium direaksikan dengan 49 gram asam sulfat menurut reaksi: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + H2(g). Tentukan pereaksi pembatas, hitung mol reaktan yang bersisa, dan hitung volume gas H2 yang dihasilkan pada keadaan STP.

Solve a stoichiometry problem involving the reaction between aluminum and sulfuric acid. Learn how to identify the limiting reactant, calculate the remaining reactant, and determine the volume of hydrogen gas produced at STP. Suitable for high school students studying chemistry.

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