Question

Use your knowledge about solubility rules and reactions to write (1) a balanced chemical equation (molecular equation), (2) a total ionic equation, and (3) a net ionic equation for the reactions below. Include the appropriate phase for each species.

G) NaOH (aq) Na2CO3(aq (k) NaOH (aq) SrCl2(aq) (i) NaOH (aq) H2SO4 (aq) (m) Na2CO3(aq) 2(aq) (n) Na2CO3(aq) H2SO4 (aq) (o) SrCl2(aq) H2SO4(aq)

Answer 1

j) NaOH (aq) + Na2CO3 (aq) --> There is no reaction that takes place.

K) 2 NaOH + SrCl2 -- > Sr(OH)2 + 2 NaCl (s) (balanced chemical equation)

      2 Na⁺ (aq) + 2 OH^- (aq) + Sr²⁺ (aq) + 2 Cl^- (aq) -----> Sr²⁺ (aq) + 2 OH^- (aq) + 2 NaCl (s) (total Ionic equation)

If we cancle the speactator ions in the above total ionic equation , we will get net ionic equation

     2 Na⁺ (aq) + 2 Cl^- (aq) -----> 2 NaCl (s)

l) 2 NaOH (aq) + H2SO4 (aq) ---- > Na₂SO₄ (aq) + 2 H₂O (l)( balanced equation)

   2 Na⁺ (aq) + 2 OH^- (aq) + 2 H⁺ (aq) + SO₄^2- (aq) -----> 2 Na⁺ + SO₄^2- (aq) + 2 H₂O (l) (total Ionic equation)

If we cancle the speactator ions in the above total ionic equation , we will get net ionic equation

2 OH^- (aq) + 2 H⁺ (aq) --> 2 H₂O (l)

m) Na2CO3 (aq) + SrCl2(aq) ---> 2 NaCl (aq) + SrCO3 (s) (Balanced equation)

    2 Na⁺ + CO₃^2- + Sr²⁺ + 2 Cl^- ----> 2 Na⁺ + 2 Cl^- + SrCO3 (s)(total Ionic equation)

If we cancle the speactator ions in the above total ionic equation , we will get net ionic equation

      CO₃^2- + Sr²⁺ ----> SrCO3 (s)

Answered first 4 sub-question in the