Question

2. How many grams of barium sulfate will form when 50.0mL of 0.125M sodium sulfate and 12.5mL of 0.60M barium chloride are mixed? Also provide the net ionic equation.

Na₂SO_{4 (aq)} + BaCl_{2 (aq)} à 2NaCl_(aq) + BaSO_4(s)

3. Write the molecular, total ionic, and net ionic equations for the following:

Aqueous Magnesium chloride reacting with Aqueous Sodium hydroxide

Aqueous Sodium hydroxide and aquous iron (III) chloride.

Solid sodium carbonate and aqueous potassium chloride

5. Finish the equation, balance it, tell what is the limiting reagent and how much starting material is left over. 53.6 lead nitrate and 26.6 of a 3.0 Potassium Sulfate

             Pb(NO_3­)_{2 (aq)} + K₂SO_{4 (aq)}   

Answer 1

Na₂SO_{4 (aq)} + BaCl_{2 (aq)} ---------------> 2NaCl_(aq) + BaSO_4(s)

no of moles of Na2So4   = molarity * volume in L

                                         = 0.125*0.05 = 0.00625 moles

no of moles of BaCl2    = molarity * volume in L

                                       = 0.6*0.0125   = 0.0075 moles

1mole of BaCl2 react with 1 mole of Na2SO4

0.0075 moles of BaCl2 react with 0.0075 moles of Na2So4

Na2So4 is limiting reactant.

1 mole of Na2So4 react with BaCl2 to gives 1 mole of BaSO4

0.00625 moles of Na2SO4 react with 0.00625 moles of BaSO4

mass of BaSO4 = no of moles * gram molar mass

                            = 0.00625*233    = 1.456g

2Na^+ (aq) +SO4^2- (aq) + Ba^2+(aq) +2Cl^- _(aq) ---------------> 2Na^+(aq) +2Cl^-_(aq) + BaSO_4(s)

SO4^2- (aq) + Ba^2+(aq) ---------------> BaSO_4(s)   net ionic equation

3. MgCl2(aq) + 2NaOH(aq) -----------------> Mg(OH)2(s) + 2NaCl(aq)

   Mg^2+ (aq) +2Cl^-(aq) + 2Na^+ (aq)+2OH^-(aq) -----------------> Mg(OH)2(s) + 2Na^+ (aq) +2Cl^-(aq)

Mg^2+ (aq) +2OH^-(aq) -----------------> Mg(OH)2(s)   net ionic equation

Na2CO3 (s) + KCl(aq) ---------------------. 2NaCl (aq) + K2CO3 (aq)