How many moles of lead (II) hydroxide (solid) can be formed when 0.0225L of 0.135M Pb (NO3) 2 solution reacts with excess sodium hydroxide?

Moles of lead (II) hydroxide (solid) forms = ?

0.0225 L of 0.135 M Pb (NO3) 2 reacts with = > sodium hydroxide

equation will read as: 0.135 M = n moles of Pb (NO3) 2 / 0.0225L, it is because: molarity = moles/liter

Apply cross multiplication above (molarity = moles / liter)

It will provide you the answer of 0.0030375 moles of Pb (NO3) 2 = > convert to Pb (OH) 2

Use the following equation where = Pb (NO3) 2 x Pb (OH) 2 / Pb (NO3) 2

= 0.0030375 Pb (NO3) 2 x 2 m Pb (OH) 2 / Pb (NO3) 2

Answer: = 0.00607 m of Pb (OH) 2