Write the net ionic equation for the aqueous solutions of manganese (II) bromide and barium sulfide.
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To determine the net ionic equation for the reaction between manganese (II) bromide (MnBr2) and barium sulfide (BaS), we first need to know the solubility of the compounds.
Manganese (II) bromide is a soluble compound, while barium sulfide is insoluble in water. Therefore, barium sulfide will form a precipitate when mixed with manganese (II) bromide.
The balanced molecular equation for the reaction is:
MnBr2(aq) + BaS(aq) → BaBr2(aq) + MnS(s)
Now, let's write the complete ionic equation by separating the soluble compounds into their respective ions:
Mn^2+(aq) + 2Br^-(aq) + Ba^2+(aq) + S^2-(aq) → Ba^2+(aq) + 2Br^-(aq) + MnS(s)
Finally, we can write the net ionic equation by eliminating the spectator ions (ions that appear on both sides of the equation and do not participate in the reaction):
Mn^2+(aq) + S^2-(aq) → MnS(s)
Therefore, the net ionic equation for the reaction between manganese (II) bromide and barium sulfide is Mn^2+(aq) + S^2-(aq) → MnS(s).
To write the net ionic equation for the reaction between manganese (II) bromide and barium sulfide, we first need to determine the formulas of each compound and then balance the equation.
The formula for manganese (II) bromide is MnBr2, indicating that it contains one manganese ion (Mn2+) and two bromide ions (Br-).
The formula for barium sulfide is BaS, indicating that it contains one barium ion (Ba2+) and one sulfide ion (S2-).
When the two compounds react, the positive ions will combine with the negative ions to form new compounds, while the bromide and sulfide ions will remain unchanged as spectator ions.
The balanced equation for the reaction between manganese (II) bromide and barium sulfide is:
MnBr2 (aq) + BaS (aq) → MnS (s) + BaBr2 (aq)
To write the net ionic equation, we remove the spectator ions, which are the bromide ions (Br-) and barium ions (Ba2+), because they appear on both sides of the equation.
Therefore, the net ionic equation for the reaction is:
Mn2+ (aq) + S2- (aq) → MnS (s)
This simplified equation shows the actual chemical change taking place, where the manganese (II) ion reacts with the sulfide ion to form solid manganese sulfide.
MnBr2(aq) + BaS(aq) = MnS(s) + BaBr2(aq)
Write the total ionic form for each. For the (s) keep it molecular form.
Mn^2+(aq) + 2Br^-(aq) + Ba^2+(aq) + S^2-(aq) = MnS(s) + Ba^2+(aq) + 2Br^-(aq)
Now cancel those ions that appear on both sides. What's left is the NET ionic equation. Post your answer if you want us to check it.