Question

Complete this net ionic equation and show your work:

Answer 1

Answer:

SO₄(aq) + Ba(aq) +  ----> BaSO₄(s)

Explanation:

(aq) will always become 2 ions

balancing the equation

Na₂SO₄(aq) + BaCl₂(aq) ----> BaSO₄ + 2NaCl

note how there is the same amount of each element on both sides except for Na and Cl. the left side has 2 times the amount of the right side. All I needed to do to balance was add the 2 coefficient to NaCl product on right side since Na and Cl are together and not in separate compounds.

now rewriting (NaCl is aq since all compounds with Na are aqueous)

barium sulfate is the solid formed

rewriting the equation: (subscripts of single elements become coefficients, polyatomic subscripts stay)

2Na(aq) + SO₄(aq) + Ba(aq) + 2Cl(aq) ----> BaSO₄(s) + 2Na(aq) + 2Cl(aq)

net ionic: removing any substance seen on both sides

SO₄(aq) + Ba(aq) +  ----> BaSO₄(s)

Answer 2

Answer : The net ionic equation is:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The balanced molecular equation will be,

$Na_2SO_4(aq)+BaCl_2(aq)\rightarrow 2NaCl(aq)+BaSO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$2Na^+(aq)+SO_4^{2-}(aq)+Ba^{2+}(aq)+2Cl^{-}(aq)\rightarrow 2Na^+(aq)+2Cl^{-}(aq)+BaSO_4(s)$

In this equation the species present are, $2Na^+\text{ and }2Cl^{-}$ are the spectator ions.

By removing the spectator ions, we get the net ionic equation.

The net ionic equation is:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$