Question

100.0 mL of 0.500 M lead(II) nitrate (Pb(NO3)2, 331 g/mol) and 100.0 mL of 0.500 M sodium sulfate (Na2SO4, 142 g/mol) are mixed together.

(I) Write the overall balanced equation. Include the states of each species.

(II) Write the overall (or total) ionic equation for the reaction. Identify any spectator ions. Include states of each species.

(III) Write the net ionic equation. Include states of each species.

(IV) Determine the limiting reactant and the theoretical yield of the non-aqueous product.

(V) Suppose this reaction was carried out in the laboratory. 0.005g of the non-aqueous product was collected. What was the percent yield?

Answer 1

(i) In balaned chemical reaction, atoms on either side of reaction are same.So, reaction is:

Pb(NO₃)₂ (aq) + Na₂SO₄(aq) PbSO₄ (s) + 2NaNO₃ (aq)

(II) Ionic equation is :

Pb²⁺(aq) + 2NO₃^- (aq) + 2Na⁺(aq) + SO₄^2-(aq) PbSO₄(s)+ 2NO₃^- (aq) + 2Na⁺(aq)

Spectator ions:  an ion that exists in the same form on both side of reaction.Thus, NO₃^- and Na⁺(aq) are spectator ions.

(III)Net ionic equation is:

Pb²⁺(aq) + SO₄^2-(aq) PbSO₄ (s)

(iV) Molarity of Pb(NO₃)₂ =0.500M

Volume of Pb(NO₃)₂ = 100 mL = 0.100L

Moles of Pb(NO₃)₂ = Molarity * Volume = 0.500M *0.1L = 0.05 moles

Molar mass of Pb(NO₃)₂ = 331 g/mol

Mass of Pb(NO₃)₂ = moles * molar mass = 0.05 moles* 331g/mol = 16.55 g

Molarity of Na₂SO₄ = 0.5M

Volume of Na₂SO_{4 = 100 mL =} 0.1L

Moles of Na₂SO₄ = 0.5M *0.1L = 0.05 moles

Molar mass of Na₂SO₄ = 142 g/mol

Mass of Na₂SO₄ = 0.05*142 = 7.1 g

From reaction 1 mol of Pb(NO₃)₂ reacts with 1 mol of Na₂SO₄

So, 0.05 mol of Pb(NO₃)₂ reacts with 0.05 mol of Na₂SO₄

So, both reacts completely.

No one is a limiting reagent.

1 mol of each reactant gives 1 mol of non-aqeous product

Thus, 0.05 mol of reactant forms 0.05 mol of non-aqueous product.

Molar mass of PbSO₄ = 303.26 g/mol

Mass of PbSO₄ = moles * molar mass = 0.05mol * 303.26 g/mol = 15.163 g

(V) Experimental yield = 0.005 g

Percent yield = (Experimental yield/ theoretical yield)*100%

= (0.005g/ 15.163g)*100% = 0.033%