Question

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of $PbO$ and 3.2 g of $HCl$? (Atomic wt. of $Pb=$ 207)

$PbO+2HCl\to PbC{l}_2+H_{2}O$

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

Answer 1

The correct option is B 0.029

$PbO+2HCl\to PbC{l}_2+H_{2}O$

Atomic mass of PbO = 207 + 16 = 223

Atomic mass of HCl = 36.5

Moles of PbO = $\frac{6.5}{223}$ = 0.029 moles

Moles of HCl = $\frac{3.2}{36.5}$ = 0.088 moles

From the above chemical equation, we see that PbO is the limiting reagent.

$\therefore$ 1 mole of PbO gives 1 mole of PbCl${}_2$

0.029 moles of PbO forms 0.029 moles of PbCl${}_2$.

Hence, the answer is option B.