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

the answer is d.) potassium