How many moles of Pb(NO3)2 are required
if 8 moles of AICI: are consumed?
3Pb(NO3)2 + 2AlCl3 + 3PbCl2 + 2A1(NO3)3

12 moles Pb(NO₄)₂ needed.

Explanation:

                3Pb(NO₃)₂ +2AlCl₃     => 3PbCl₂ + 2Al(NO₃)₃

Given =>   ? moles      8 moles

from reaction stoichiometry, 2 moles AlCl₃ requires 3 moles Pb(NO₄)₂ then 8 moles AlCl₃ requires 3/2(8) moles of the Pb(NO₄)₂ => 12 moles Pb(NO₄)₂ needed.

6.022×10²¹

Explanation:

First calculate the number of moles

n=m/GFM

n=.018/18

n=0.01moles

Then multiply the number of moles by the molar mass

0.01×6.022×10²³ =6.022×10²¹