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. (a)`0.333`
B. (b)`0.011`
C. (c )`0.029`
D. (d)`0.044`

Answer

C

Explanation

`PbO+2HCl rarr PbCl_(2) rarr PbCl_(2)+H_(2)O`
`n_(PbO)=6.5/223=0.029`
`n_(HCl)=3.2/36.5=0.087`
`{:(PbO " gives PbCl"_(2),|,"HCL gives" ,PbCl_(2),),("Moles " 0.029,0.087,0.048,,),(,impliesn_(PbCl_(2)),=0.029,,):}`