How many moles of lead (II) choride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?

Options

A. 0.044
B. 0.333
C. 0.011
D. 0.029

Explanation

To solve the problem of how many moles of lead (II) chloride (PbCl2) will be formed from the reaction between 6.5 g of lead oxide (PbO) and 3.2 g of hydrochloric acid (HCl), we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction between lead oxide and hydrochloric acid can be represented by the following balanced equation:

\[ \text{PbO} + 2 \text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the Molar Masses
- **Molar mass of PbO**:
- Lead (Pb) = 207 g/mol
- Oxygen (O) = 16 g/mol
- Molar mass of PbO = 207 + 16 = 223 g/mol

- **Molar mass of HCl**:
- Hydrogen (H) = 1 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

### Step 3: Calculate the Number of Moles of Each Reactant
- **Moles of PbO**:
\[ \text{Moles of PbO} = \frac{\text{mass}}{\text{molar mass}} = \frac{6.5 \text{ g}}{223 \text{ g/mol}} \approx 0.029 \text{ moles} \]

- **Moles of HCl**:
\[ \text{Moles of HCl} = \frac{3.2 \text{ g}}{36.5 \text{ g/mol}} \approx 0.088 \text{ moles} \]

### Step 4: Determine the Limiting Reactant
From the balanced equation, we see that:
- 1 mole of PbO reacts with 2 moles of HCl.

Now, we need to check how many moles of HCl are required to react with 0.029 moles of PbO:
\[ \text{Required moles of HCl} = 0.029 \text{ moles PbO} \times 2 = 0.058 \text{ moles HCl} \]

Since we have 0.088 moles of HCl available, HCl is in excess, and PbO is the limiting reactant.

### Step 5: Calculate the Moles of PbCl2 Formed
According to the balanced equation, 1 mole of PbO produces 1 mole of PbCl2. Therefore, the moles of PbCl2 produced will be equal to the moles of PbO reacted:
\[ \text{Moles of PbCl2} = 0.029 \text{ moles PbO} = 0.029 \text{ moles PbCl2} \]

### Final Answer
The number of moles of lead (II) chloride (PbCl2) formed is **0.029 moles**.

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