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How many moles of lead (II) chloride will be formed from a reaction between 6.5 g6.5\text{ g} of PbO\text{PbO} and 3.2 g3.2\text{ g} of HCl\text{HCl}?

A

0.0440.044

B

0.3330.333

C

0.0110.011

D

0.0290.029

Step-by-Step Solution

The balanced chemical equation is: PbO+2HClPbCl2+H2O\text{PbO} + 2\text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O}

Molar mass of PbO=207.2+16=223.2 g/mol\text{PbO} = 207.2 + 16 = 223.2\text{ g/mol}. Moles of PbO=6.5 g223.2 g/mol0.0291 mol\text{PbO} = \frac{6.5\text{ g}}{223.2\text{ g/mol}} \approx 0.0291\text{ mol}.

Molar mass of HCl=1+35.5=36.5 g/mol\text{HCl} = 1 + 35.5 = 36.5\text{ g/mol}. Moles of HCl=3.2 g36.5 g/mol0.0876 mol\text{HCl} = \frac{3.2\text{ g}}{36.5\text{ g/mol}} \approx 0.0876\text{ mol}.

From the balanced equation, 1 mole1\text{ mole} of PbO\text{PbO} reacts with 2 moles2\text{ moles} of HCl\text{HCl}. So, 0.0291 moles0.0291\text{ moles} of PbO\text{PbO} will require 0.0291×2=0.0582 moles0.0291 \times 2 = 0.0582\text{ moles} of HCl\text{HCl}. Since 0.0876 moles0.0876\text{ moles} of HCl\text{HCl} are available, which is more than the required amount (0.0582 moles0.0582\text{ moles}), PbO\text{PbO} is the limiting reagent and HCl\text{HCl} is in excess.

The amount of product formed depends completely on the limiting reagent. 1 mole1\text{ mole} of PbO\text{PbO} produces 1 mole1\text{ mole} of PbCl2\text{PbCl}_2. Therefore, 0.029 moles0.029\text{ moles} of PbO\text{PbO} will produce 0.029 moles0.029\text{ moles} of PbCl2\text{PbCl}_2.

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