9.70g of sodium sulphide. Na2S is added to 550 mL of 0.200 mol/L lead (II) nitrate solution, Pb(NO3)2. Calculate the maximum mass of precipitate that can form.

Question

I need all the steps to find the answer, preferably with the G.R.A.S.S method please.

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Answer 1

I'm not familiar with the G.R.A.S.S. method, so my apologies for that.^**

(1) Write the correctly balanced equation for the reaction:

Na₂S(aq) + Pb(NO₃)₂(aq) ==> 2NaNO₃(aq) + PbS(s) ... balanced equation

(2) Find the limiting reactant:

Easy way is to find moles of each reactant and dived by the corresponding coefficient in the equation...

For Na₂S: 9.70 g x 1 mol / 78.0 g = 0.124 mols Na₂S (÷1->0.124)

For Pb(NO₃)₂: 550 ml x 1 L/1000 ml x 0.200 mol/L = 0.110 mols Pb(NO₃)₂ (÷1->0.11)

Since 0.11 is less than 0.124, Pb(NO₃)₂ is limiting and will dictate how much PbS is formed.

(3) Use moles Pb(NO₃)₂ to calculate grams of PbS formed:

0.110 mols Pb(NO₃)₂ x 1 mol PbS / mol Pb(NO₃)₂ x 239 g / mol Pb(NO₃)₂ = 26.3 g Pb(NO₃)₂

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** After submitting the answer, I looked up the GRASS method and found it to be some inane process which isn't really necessary. It's one of these "new" inventions for teaching. Not sure what's wrong with the old way of doing it. It stands for Given, Required, Application, Solution, Statement.