For the reaction pb(no3)2 + 2ki → pbi2 + 2kno3, how many moles of lead iodide are produced from 246.2 g of potassium iodide?

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

Ibelieve the answer is d