00:01
Good day in this question we are asked to solve for the mass of precipitate that will form.
00:05
So with a given balance equation here, we have sodium sulfate, aqueous plus lead nitrate, aqueous, that will yield to lead sulfate, solid, and sodium, nitrate, equs, that will yield to lead sulfate, solid, solid, and sodium, and sodium, nitrate, aqub, so the precipitate here is the solid in the reactant.
00:49
Therefore, this is our precipitate.
00:56
So next we identify the limiting reactant by solving the moles of each reactant.
01:02
We have mole of sodium sulfate is equals to the malarity times the volume in liters.
01:09
Given malarity is 0 .10 moles per liter times the volume of 25ml...