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

Options

A. 0.011
B. 0.029
C. 0.44
D. 0.33

Explanation

To determine 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 as:
\[ \text{PbO} + 2 \text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass 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**:
\[ \text{Moles of PbO} = \frac{\text{mass}}{\text{molar mass}} = \frac{6.5 \text{ g}}{223 \text{ g/mol}} \approx 0.0292 \text{ moles} \]

- **Moles of HCl**:
\[ \text{Moles of HCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.2 \text{ g}}{36.5 \text{ g/mol}} \approx 0.0877 \text{ moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that 1 mole of PbO reacts with 2 moles of HCl. Therefore, we need to find out how many moles of HCl are required for the available moles of PbO:
- Required moles of HCl for 0.0292 moles of PbO:
\[ 0.0292 \text{ moles PbO} \times 2 = 0.0584 \text{ moles HCl} \]

Since we have 0.0877 moles of HCl available, HCl is in excess, and PbO is the limiting reagent.

### Step 5: Calculate the moles of lead (II) chloride 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:
\[ \text{Moles of PbCl}_2 = \text{Moles of PbO} = 0.0292 \text{ moles} \]

### Conclusion
The number of moles of lead (II) chloride formed from the reaction is approximately **0.0292 moles**.

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