How many milliliters of 0.10 m pb(no3)2 are required to react with 75 ml of 0.20 m nai? pb(no3)2 + 2 nai --> pbi2 + 2 nano3?

The Ml of 0.10 Pb(NO3)2 that are required to react with 75 Ml of 0.20M Nai is 75 ml

calculation

Pb(NO3)2 +2Nai → Pbi2 + 2NaNo3

find the number of moles of Nai used

moles= molarity x volume in liters

volume in liters = 75/1000=0.075 moles

mole =0.20 x0.075 = 0.015 moles

by use of mole ratio between Pb(NO3)2 to Nai which is 1 :2 the moles of Pb(NO3)2 = 0.015 x1/2 =7.5 x10^-3 moles

volume (ml) )= moles/molarity x 1000

=volume = (7.5 x10^-3) / 0.10 = 75 Ml

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.