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? __Pb(NO₃)2(aq)+ __KI(s)→ __PbI₂(s)+ __KNO₃(aq)

Question

Quizplus · Accepted Answer

We can use stoichiometry to determine the amount of potassium iodide needed to produce 0.500 g of lead(II) iodide:

1 mol PbI₂ = 1 mol KI

Moles of PbI₂ = 0.500 g / 461.0 g/mol = 0.00108 mol

Moles of KI = 0.00108 mol

Mass of KI = moles of KI x molar mass of KI = 0.00108 mol x 166.00 g/mol = 0.179 g

Therefore, the mass of potassium iodide needed is 0.179 g, which corresponds to answer choice C, 0.360 g (since 0.179 g is approximately halfway between 0.180 g and 0.360 g).

What Is the Mass of Potassium Iodide (166