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

Options

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

Answer

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 can follow these steps:

### Step 1: Write the balanced chemical equation.
The reaction between lead oxide and hydrochloric acid can be represented as:
\[ \text{PbO} + 2 \text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the molar masses of the reactants.
- **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**:
\[ \text{Moles of HCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.2 \text{ g}}{36.5 \text{ g/mol}} \approx 0.088 \text{ moles} \]

### Step 4: Determine the limiting reagent.
From the balanced equation, we see that 1 mole of PbO reacts with 2 moles of HCl. Therefore:
- The moles of HCl required for 0.029 moles of PbO:
\[ \text{Required HCl} = 0.029 \text{ moles PbO} \times 2 = 0.058 \text{ moles HCl} \]

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

### Step 5: Calculate the moles of PbCl2 produced.
According to the stoichiometry of the reaction, 1 mole of PbO produces 1 mole of PbCl2. Therefore:
\[ \text{Moles of PbCl2} = \text{Moles of PbO} = 0.029 \text{ moles} \]

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

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