Respuesta :
Answer: The equations are given below.
Explanation:
To write the net ionic equations, we do not include spectator ions in the equations.
Spectator ions are defined as the ions which do not participate in a chemical reactions. These ions exists in the same form on both the sides of the reaction.
Equations for the given reactions is given as:
- a.) Copper(II) sulfate, [tex]CuSO_4[/tex], and potassium hydroxide, KOH
Ionic equation follows:
[tex]Cu^{2+}(aq.)+SO_4^{2-}(aq.)+2K^+(aq.)+2OH^-(aq.)\rightarrow Cu(OH)_2(s)+2K^+(aq.)+SO_4^{2-}(aq.)[/tex]
Removing the spectator ions, we get the net ionic equation, which is:
[tex]Cu^{2+}(aq.)+2OH^-(aq.)\rightarrow Cu(OH)_2(s)[/tex]
- b.) Lithium carbonate, [tex]Li_2CO_3[/tex], and aluminum nitrate, [tex]Al(NO_3)_3[/tex]
Ionic equation follows:
[tex]6Li^+(aq.)+6CO_3^{2-}(aq.)+2Al^{3+}(aq.)+6NO_3^-(aq.)\rightarrow 6Li^+(aq.)+6NO_3^-(aq.)+Al_2(CO_3)_3(s)[/tex]
Removing the spectator ions, we get the net ionic equation, which is:
[tex]6CO_3^{2-}(aq.)+2Al^{3+}(aq.)\rightarrow Al_2(CO_3)_3(s)[/tex]
- c.) Sodium phosphate, [tex]Na_3PO_4[/tex], and barium chloride, [tex]BaCl_2[/tex]
Ionic equation follows:
[tex]6Na^+(aq.)+2PO_4^{3-}(aq.)+3Ba^{2+}(aq.)+6Cl^-(aq.)\rightarrow Ba_3(PO_4)_2(s)+6Na^+(aq.)+6Cl^-(aq.)[/tex]
Removing the spectator ions, we get the net ionic equation, which is:
[tex]3Ba^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Ba_3(PO_4)_2(s)[/tex]
Hence, the equations are given above.
The balanced net ionic equation for the reactions are:
A. The net ionic equation for the reaction between copper (II) sulfate, CuSO₄, and potassium hydroxide, KOH is:
Cu²⁺(aq) + 2OH¯(aq) —> Cu(OH)₂(s)
B. The net ionic equation for the reaction between lithium carbonate, Li₂CO₃, and aluminum nitrate, Al(NO₃)₃ is:
3CO₃²¯(aq) + 2Al³⁺(aq) —> Al₂(CO₃)₃(s)
C. The net ionic equation for the reaction between sodium phosphate, Na₃PO₄, and barium chloride, BaCl₂ is:
2PO₄³¯(aq) + 3Ba²⁺(aq) —> Ba₃(PO₄)₂(s)
A. Determination of the net ionic equation for the reaction between copper (II) sulfate, CuSO₄, and potassium hydroxide, KOH
In solution, CuSO₄ and KOH will dissociate and react as follow:
CuSO₄(aq) —> Cu²⁺(aq) + SO₄²¯(aq)
KOH(aq) —> K⁺(aq) + OH¯(aq)
CuSO₄(aq) + KOH(aq) —>
Cu²⁺(aq) + SO₄²¯(aq) + K⁺(aq) + OH¯(aq) —> Cu(OH)₂(s) + SO₄²¯(aq) + K⁺(aq)
Cancel the spectator ions (i.e SO₄²¯ and K⁺) and balance the equation to obtain the net ionic equation as shown below:
Cu²⁺(aq) + 2OH¯(aq) —> Cu(OH)₂(s)
B. Determination of the net ionic equation for the reaction between lithium carbonate, Li₂CO₃, and aluminum nitrate, Al(NO₃)₃
In solution, the reaction will proceed as follow:
Li₂CO₃(aq) + Al(NO₃)₃(aq) —>
2Li⁺(aq) + CO₃²¯(aq) + Al³⁺(aq) + 3NO₃¯(aq) —> Al₂(CO₃)₃(s) + 2Li⁺(aq) + 3NO₃¯(aq)
Cancel the spectator ions (i.e Li⁺ and NO₃¯) and balance the equation to obtain the net ionic equation as shown below:
3CO₃²¯(aq) + 2Al³⁺(aq) —> Al₂(CO₃)₃(s)
C. Determination of the net ionic equation for the reaction between sodium phosphate, Na₃PO₄, and barium chloride, BaCl₂
Na₃PO₄(aq) + BaCl₂(aq) —>
3Na⁺(aq) + PO₄³¯(aq) + Ba²⁺(aq) + 2Cl¯(aq) —> Ba₃(PO₄)₂(s) + 3Na⁺(aq) + 2Cl¯(aq)
Cancel the spectator ions (Na⁺ and Cl¯) and balance the equation to obtain the net ionic equation as shown below:
2PO₄³¯(aq) + 3Ba²⁺(aq) —> Ba₃(PO₄)₂(s)
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