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Balance the given molecular equations in aqueous solution with coefficients and phases for each species. If you determine that a species should have a coefficient of 1, you must indicate this with the "1" label. Do not leave the coefficient space blank for any species in the balanced equation.

1 2 3 4 5 6 s aq

K2SO4 ( )+ Pb(NO3)2 ( ) → KNO3 ( )+ PbSO4 ( )
Ca(C2H3O2)2 ( )+ (NH4)2CO3 ( ) → CaCO3()+ NH4C2H3O2()
Li3PO4( )+ BaCl2 ( ) → LiCl ( ) + Ba3(PO4)2 ( )
MgBr2( ) + NaOH ( ) → Mg(OH)2 ( ) + NaBr ( )

Respuesta :

sebassandin

Answer:

- [tex]1K_2SO_4 (aq)+ 1Pb(NO_3)_2 (aq) \rightarrow 2KNO_3 (aq)+ 1PbSO_4 (s)[/tex]

- [tex]1Ca(C_2H_3O_2)_2 (aq)+ 1(NH_4)_2CO_3 (aq) \rightarrow 1CaCO_3(s)+ 2NH_4C_2H_3O_2(aq)[/tex]

- [tex]2Li_3PO_4(aq)+3 BaCl_2 (aq) \rightarrow 6LiCl (aq) + Ba_3(PO_4)_2 (s)[/tex]

- [tex]1MgBr_2(aq) + 2NaOH (aq) \rightarrow 1Mg(OH)_2 (s) + 2NaBr (aq)[/tex]

Explanation:

Hello,

In this case, we must make equal the number of atoms of each element at both reactants and products as shown below:

(a) Lead (II) sulfate precipitates out.

[tex]K_2SO_4 (aq)+ Pb(NO_3)_2 (aq) \rightarrow KNO_3 (aq)+ PbSO_4 (s)\\\\1K_2SO_4 (aq)+ 1Pb(NO_3)_2 (aq) \rightarrow 2KNO_3 (aq)+ 1PbSO_4 (s)[/tex]

(b) Calcium carbonate precipitates out.

[tex]Ca(C_2H_3O_2)_2 (aq)+ (NH_4)_2CO_3 (aq) \rightarrow CaCO_3(s)+ NH_4C_2H_3O_2(aq)\\\\1Ca(C_2H_3O_2)_2 (aq)+ 1(NH_4)_2CO_3 (aq) \rightarrow 1CaCO_3(s)+ 2NH_4C_2H_3O_2(aq)[/tex]

(c) Barium phosphate precipitates out.

[tex]Li_3PO_4(aq)+ BaCl_2 (aq) \rightarrow LiCl (aq) + Ba_3(PO_4)_2 (s)\\\\2Li_3PO_4(aq)+3 BaCl_2 (aq) \rightarrow 6LiCl (aq) +1Ba_3(PO_4)_2 (s)[/tex]

(d) Magnesium hydroxide precipitates out.

[tex]MgBr_2(aq) + NaOH (aq) \rightarrow Mg(OH)_2 (s) + NaBr (aq)\\\\1MgBr_2(aq) + 2NaOH (aq) \rightarrow 1Mg(OH)_2 (s) + 2NaBr (aq)[/tex]

Best regards.

onyebuchinnaji

The balanced chemical equation is written as;

[tex]K_2SO_4_{(aq)} \ + \ Pb(NO_3)_2_{(aq)} \ --> \ 2KNO_3_{(aq)} \ + \ PbSO_4 _{(s)} \\\\Ca(C_2H_3O_2)_2_{(aq)} + (NH_4)_2CO_3_{(aq)} \ -> \ CaCO_3_{(s)} + 2NH_4C_2H_3O_2 _{(aq)} \\\\2Li_3PO_4 _{(aq)} \ + \ 3BaCl_2_{(aq)} \ -->\ 6LiCl_{(aq)} \ + \ Ba_3(PO_4)_2 _{(s)} \\\\MgBr_2_{(aq)} \ + \ 2NaOH_{(aq)} \ --> \ Mg(OH)_2_{(s)} \ + \ 2NBr_{(aq)}[/tex]

The balanced chemical equation for the given compounds is written as follows;

  • the number of each element in the reactant side must equal the product side
  • aq ---- aqueous solution
  • s ---- solid precipitate

[tex]K_2SO_4_{(aq)} \ + \ Pb(NO_3)_2_{(aq)} \ --> \ 2KNO_3_{(aq)} \ + \ PbSO_4 _{(s)} \\\\Ca(C_2H_3O_2)_2_{(aq)} + (NH_4)_2CO_3_{(aq)} \ -> \ CaCO_3_{(s)} + 2NH_4C_2H_3O_2 _{(aq)} \\\\2Li_3PO_4 _{(aq)} \ + \ 3BaCl_2_{(aq)} \ -->\ 6LiCl_{(aq)} \ + \ Ba_3(PO_4)_2 _{(s)} \\\\MgBr_2_{(aq)} \ + \ 2NaOH_{(aq)} \ --> \ Mg(OH)_2_{(s)} \ + \ 2NBr_{(aq)}[/tex]

Thus, the balanced chemical equation of the given various reaction is obtained by balancing each element in the reactant side and product side.

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