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

Options

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

Answer

D

Explanation

`{:(Pbo,+,2HCl,to,PbCl_(2),+,H_(2)O),(207+16,,2xx36.5,,(1 "mole"),,),(=223,,=73,,,,),((1 "mole"),,(2 "mole"),,,,):}`
Mole of PbO `=(6.5)/(223)=0.029`
Mole of HCl `=(3.2)/(36.5)=0.087`
Since 1 mole of PbO reacts with 2 moles of HCl , thus in this reaction PbO is the limiting reagent.
Hence , 1 mole of PbO forms =1 mole of `PbCl_(2)`
0.029 mole of PbO will form =0.029 mole of `PbCl_(2)`