2. How many grams of barium sulfate will form when 50.0mL of 0.125M sodium sulfa
ID: 554519 • Letter: 2
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.
Na2SO4 (aq) + BaCl2 (aq) à 2NaCl(aq) + BaSO4(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(NO3)2 (aq) + K2SO4 (aq)
Explanation / Answer
Na2SO4 (aq) + BaCl2 (aq) ---------------> 2NaCl(aq) + BaSO4(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) + BaSO4(s)
SO4^2- (aq) + Ba^2+(aq) ---------------> BaSO4(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)
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