What is the mass of potassium iodide (166.00 g/mol) that yields 0.500 g of lead (ii) iodide (461.0 g/mol) precipitate? (you must balance the equation first) __pb (no3) 2 (aq) + __ki (s) → __pbi2 (s) + __kno3 (aq) 0.0900 g 2.78 g 0.694 g 0.180 g 0.360 g?

the mass of potassium iodide is 0.360 g

calculation

step 1: write the balanced molecular equation

Pb (NO3) 2 (aq) + 2 Ki (s) → Pbi2 (s) + 2KNO3

step 2; calculate the moles of Pbi2

moles = mass / molar mass

= 0.500 g / 461.0 g/mol = 0.0011 moles

step 3: use the mole ratio to calculate the moles of Ki

Ki: Pbi2 is 2:1 therefore the moles of ki = 0.0011 x 2/1 = 0.0022 moles

step 4 : find the mass of Ki

mass = moles x molar mass

=0.0022 moles x 166g/mol = 0.365 g which is approximate 0.360 g