Pb (NO3) 2 + 2Kl - --> 2KNO3 + PbI2

How many grams of lead (ll) nitrate is needed to completely react 25g or potassium iodide?

Mass = 24.84 g

Explanation:

Given dа ta:

Mass of potassium iodide = 25 g

Mass of lead nitrate required = ?

Solution:

Chemical equation:

Pb (NO₃) ₂ + 2KI → 2KNO₃ + PbI₂

Number of moles of potassium iodide:

Number of moles = mass / molar mass

Number of moles = 25 g / 166 g/mol

Number of moles = 0.15 mol

Now we will compare the moles of KI and Pb (NO₃) ₂.

KI : Pb (NO₃) ₂

2 : 1

0.15 : 1/2*0.15 = 0.075 mol

Mass of lead nitrate:

Mass = number of moles * molar mass

Mass = 0.075 mol * 331.2 g/mol

Mass = 24.84 g