43. write balanced complete ionic and net ionic equations for each reaction:\na. hcl(aq) + lioh(aq) → h₂o(l)…

43. write balanced complete ionic and net ionic equations for each reaction:\na. hcl(aq) + lioh(aq) → h₂o(l) + licl(aq)\nb. mgs(aq) + cucl₂(aq) → cus(s) + mgcl₂(aq)\nc. naoh(aq) + hno₃(aq) → h₂o(l) + nano₃(aq)\nd. na₃po₄(aq) + nicl₂(aq) → ni₃(po₄)₂(s) + nacl(aq)

43. write balanced complete ionic and net ionic equations for each reaction:\na. hcl(aq) + lioh(aq) → h₂o(l) + licl(aq)\nb. mgs(aq) + cucl₂(aq) → cus(s) + mgcl₂(aq)\nc. naoh(aq) + hno₃(aq) → h₂o(l) + nano₃(aq)\nd. na₃po₄(aq) + nicl₂(aq) → ni₃(po₄)₂(s) + nacl(aq)

Answer

Explanation:

Step1: Write complete ionic equations

a.

$HCl(aq)$ dissociates into $H^+(aq)+Cl^-(aq)$, $LiOH(aq)$ into $Li^+(aq) + OH^-(aq)$, $LiCl(aq)$ into $Li^+(aq)+Cl^-(aq)$. The complete ionic equation is $H^+(aq)+Cl^-(aq)+Li^+(aq)+OH^-(aq)\rightarrow H_2O(l)+Li^+(aq)+Cl^-(aq)$.

b.

$MgS(aq)$ dissociates into $Mg^{2 +}(aq)+S^{2-}(aq)$, $CuCl_2(aq)$ into $Cu^{2+}(aq)+2Cl^-(aq)$, $MgCl_2(aq)$ into $Mg^{2+}(aq)+2Cl^-(aq)$. The complete ionic equation is $Mg^{2 +}(aq)+S^{2-}(aq)+Cu^{2+}(aq)+2Cl^-(aq)\rightarrow CuS(s)+Mg^{2+}(aq)+2Cl^-(aq)$.

c.

$NaOH(aq)$ dissociates into $Na^+(aq)+OH^-(aq)$, $HNO_3(aq)$ into $H^+(aq)+NO_3^-(aq)$, $NaNO_3(aq)$ into $Na^+(aq)+NO_3^-(aq)$. The complete ionic equation is $Na^+(aq)+OH^-(aq)+H^+(aq)+NO_3^-(aq)\rightarrow H_2O(l)+Na^+(aq)+NO_3^-(aq)$.

d.

$Na_3PO_4(aq)$ dissociates into $3Na^+(aq)+PO_4^{3 -}(aq)$, $NiCl_2(aq)$ into $Ni^{2+}(aq)+2Cl^-(aq)$, $NaCl(aq)$ into $Na^+(aq)+Cl^-(aq)$. The complete ionic equation is $3Na^+(aq)+PO_4^{3 -}(aq)+Ni^{2+}(aq)+2Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+Na^+(aq)+Cl^-(aq)$. Multiply through by 2 to balance the phosphate and nickel ions: $6Na^+(aq)+2PO_4^{3 -}(aq)+3Ni^{2+}(aq)+6Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+6Na^+(aq)+6Cl^-(aq)$.

Step2: Write net - ionic equations

a.

Cancel out the spectator ions ($Li^+$ and $Cl^-$). The net ionic equation is $H^+(aq)+OH^-(aq)\rightarrow H_2O(l)$.

b.

Cancel out the spectator ions ($Mg^{2+}$ and $Cl^-$). The net ionic equation is $S^{2-}(aq)+Cu^{2+}(aq)\rightarrow CuS(s)$.

c.

Cancel out the spectator ions ($Na^+$ and $NO_3^-$). The net ionic equation is $H^+(aq)+OH^-(aq)\rightarrow H_2O(l)$.

d.

Cancel out the spectator ions ($Na^+$ and $Cl^-$). The net ionic equation is $2PO_4^{3 -}(aq)+3Ni^{2+}(aq)\rightarrow Ni_3(PO_4)_2(s)$.

Answer:

a. Complete ionic: $H^+(aq)+Cl^-(aq)+Li^+(aq)+OH^-(aq)\rightarrow H_2O(l)+Li^+(aq)+Cl^-(aq)$ Net ionic: $H^+(aq)+OH^-(aq)\rightarrow H_2O(l)$ b. Complete ionic: $Mg^{2 +}(aq)+S^{2-}(aq)+Cu^{2+}(aq)+2Cl^-(aq)\rightarrow CuS(s)+Mg^{2+}(aq)+2Cl^-(aq)$ Net ionic: $S^{2-}(aq)+Cu^{2+}(aq)\rightarrow CuS(s)$ c. Complete ionic: $Na^+(aq)+OH^-(aq)+H^+(aq)+NO_3^-(aq)\rightarrow H_2O(l)+Na^+(aq)+NO_3^-(aq)$ Net ionic: $H^+(aq)+OH^-(aq)\rightarrow H_2O(l)$ d. Complete ionic: $6Na^+(aq)+2PO_4^{3 -}(aq)+3Ni^{2+}(aq)+6Cl^-(aq)\rightarrow Ni_3(PO_4)_2(s)+6Na^+(aq)+6Cl^-(aq)$ Net ionic: $2PO_4^{3 -}(aq)+3Ni^{2+}(aq)\rightarrow Ni_3(PO_4)_2(s)$