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?

Question

A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is:
(Given atomic masses of A=64; B=40; C=32 u)

Answer 1

To determine the number of moles of lead (II) chloride (PbCl₂) formed, we need to consider the balanced chemical equation for the reaction between lead(II) oxide (PbO) and hydrochloric acid (HCl):

\mathrm{PbO(s) + 2HCl(aq) \rightarrow PbCl_2(s) + H_2O(l)}

The balanced equation shows that 1 mole of PbO reacts with 2 moles of HCl to form 1 mole of PbCl₂.

Step 1: Calculate the number of moles of each reactant

The molar mass of PbO (Lead(II) Oxide) is calculated as follows:

Lead (Pb) = 207.2 g/mol
Oxygen (O) = 16.0 g/mol
Molar mass of PbO = 207.2 g/mol + 16.0 g/mol = 223.2 g/mol

Number of moles of PbO:

\text{moles of PbO} = \frac{6.5 \text{ g}}{223.2 \text{ g/mol}} = 0.029 \text{ moles}

The molar mass of HCl is:

Hydrogen (H) = 1.0 g/mol
Chlorine (Cl) = 35.5 g/mol
Molar mass of HCl = 1.0 g/mol + 35.5 g/mol = 36.5 g/mol

Number of moles of HCl:

\text{moles of HCl} = \frac{3.2 \text{ g}}{36.5 \text{ g/mol}} = 0.0877 \text{ moles}

Step 2: Determine the limiting reactant

According to the reaction, 1 mole of PbO requires 2 moles of HCl. Therefore, 0.029 moles of PbO would require:

0.029 \times 2 = 0.058 \text{ moles of HCl}

Since we only have 0.0877 moles of HCl, we actually have more than enough HCl to react with all the PbO. This means HCl is in excess, and PbO is the limiting reactant.

Step 3: Calculate the moles of PbCl₂ formed from the limiting reactant

Since PbO is the limiting reactant, the number of moles of PbCl₂ formed will be equal to the number of moles of PbO that reacts:

\text{moles of PbCl}_2 = 0.029 \text{ moles}

The correct answer is therefore 0.029 moles.

This corresponds to the given option: 0.029

Answer 2