Question

Pb(NO3)2 + 2Kl ——-> 2KNO3 + PbI2

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

Answer 1

Mass = 24.84 g

Step-by-step explanation:

Given data:

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