Question

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

[Given: atomic mass of Pb = 208 u]

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

Answer 1

Answer: (B)

$\underset{\underset{\underset{0.029\text{mol}}{\frac{6.5}{224}\text{mol}}}{}}{\mathrm{PbO}}+\underset{\underset{\underset{0.087\text{mol}}{\frac{3.2}{36.5}\text{mol}}}{}}{2\text{HCl}}\to {\text{PbCl}}_2+{\text{H}}_2\text{O}$

Formation of moles of lead (II) chloride depends upon the number of moles of PbO which acts as a limiting factor here. So, number of moles of $PbC{l}_2$ formed will be equal to the number of moles of PbO i.e. 0.029.