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

Options

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

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 can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction between lead (II) oxide and hydrochloric acid is:
\[ \text{PbO} + 2 \text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the moles of PbO
To find the number of moles of PbO, we use the formula:
\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of PbO (Lead (II) oxide) is approximately 223 g/mol.

Using the given mass:
\[ \text{Moles of PbO} = \frac{6.5 \text{ g}}{223 \text{ g/mol}} \approx 0.0292 \text{ moles} \]

### Step 3: Calculate the moles of HCl
Next, we calculate the moles of HCl using its molar mass, which is approximately 36.5 g/mol:
\[ \text{Moles of HCl} = \frac{3.2 \text{ g}}{36.5 \text{ g/mol}} \approx 0.0877 \text{ moles} \]

### Step 4: Determine the limiting reactant
From the balanced equation, we see that 1 mole of PbO reacts with 2 moles of HCl. Therefore, to react with 0.0292 moles of PbO, we would need:
\[ 0.0292 \text{ moles PbO} \times 2 = 0.0584 \text{ moles HCl} \]

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

### Step 5: Calculate the moles of PbCl2 produced
According to the balanced equation, 1 mole of PbO produces 1 mole of PbCl2. Thus, the moles of PbCl2 produced will be equal to the moles of PbO reacted:
\[ \text{Moles of PbCl2} = 0.0292 \text{ moles} \]

### Final Answer
The number of moles of lead (II) chloride formed is approximately **0.0292 moles**.

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