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

Question

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

Answer 1

Always write the balanced equation:

Pb(NO₃)₂(aq) + 2NaOH(aq) ==> Pb(OH)₂(s) + 2NaNO₃(aq)

Note the mole ratio between Pb(NO₃)₂ and Pb(OH)₂ is 1 : 1. So, if we can find the moles of Pb(NO₃)₂ used, we will also know the moles of Pb(OH)₂ formed.

moles Pb(NO₃)₂ present = 0.0225 L x 0.135 mol/L = 0.0030375 moles

moles Pb(OH)₂ formed = 0.00304 moles (answer to 3 sig. figs.)