Question

For the reaction Pb(NO3)2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 246.2 g of potassium iodide?

Answer 1

m(KI)=246.2 g
M(KI)=166.0 g/mol

n(KI)=m(KI)/M(KI)
n(KI)=246.2/166.0≈1.48 mol

2 mol KI - 1 mol  PbI₂
1.48 mol KI - x mol PbI₂

x=1.48*1/2=0.74 mol