Identify the spectator ions in the following complete ionic equation. 2K+(aq)+S2−(aq)+Pb2+(aq)+2NO−3...
The spectators are the ions that appear on both sides of the arrow in the complete ionic equation. In this case they are K{+} and NO3{-} ions.
Identify the spectator ions in the following complete ionic equation.?
Ba^2(aq ) + 21^-(aq) + 2 Na^+(aq ) + SO4)^-2 (aq)--> BaSO4(s) + 21^-(aq) + Na^+(aq) Na+1 & I-1 are the spectators yoiur net ionic is only Ba^2(aq ) & (SO4)^-2, (aq) --> BaSO4(s)
Identify the spectator ions in the following reaction.?
C Spectator ions are the ones that don't get used in the reaction to form a specific product but instead remains in the solution in the form of ions... For this, you have to see the reaction states, such as solid, liquids and gas products on the either sides can't be spectator ions.. It is the aqueous ions that are spectator and remain as such on both sides and there can be eliminated... In your case since CaCO3 is forming as a solid,,, any ions forming that on left sides can't be included as spectors. Therefore Ca2+ and CO3 are out.. THe rest is spectator.. Hope that helps,
Complete each of the following as a net ionic equation. If no reaction occurs, so state. a) Ba2++2I−+2K++CO2−3→ b) Ba2++S2−+2Na++SO2−4→?
a) Ba{2+} + 2 I{-} + 2 K{+} + CO3{2-} → BaCO3(s) + 2 KI b) Ba{2+} +S{2-} + 2 Na{+} + SO4{2-} → BaSO4(s) + Na2S
Complete the following precipitation reactions?
(a) CaCl2(aq) + Cs3PO4(aq) Molecular (chemical) 3CaCl2(aq) + 2Cs3PO4(aq) --> 6CsCl(s) + Ca3(PO4)(aq) Total ionic (complete ionic) 3Ca2+ + 6Cl- + 6Cs+ + 2[PO4]3- --> 6CsCl(s) + 3Ca2+ + 2[PO4]3- Net Ionic Cl- + Cs --> CsCl(s) (b) Na2S(aq) + ZnSO4(aq) "Molecular equation" (Chemical) Na2S(aq) + ZnSO4(aq) --> Na2SO4 + ZnS Total (complete) ionic 2Na+ + S2- + Zn2+ + [SO4]2- --> ZnS(s) + 2Na+ + Zn2 Net Ionic S2- + Zn2+ --> ZnS(s)
Spectator ions in an ionic equation?
Net ionic equation: Ba++(aq) + SO4=(aq) ------> BaSO4(s) so that leaves Na+ and I- as the spectator ions.
How do you find spectator ions from an equation?
Let's do an example.Consider the following equation:Ag(NO3) (aq) + NaCl (aq) -> AgCl (s) + NaNO3 (aq) Now, we can see that some compounds are in aqueous form, and so they disassociate in water, giving us the complete ionic equation:Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) -> AgCl (s) + Na+ (aq) + NO3- (aq)Now, you can see that some of the reactants are exactly the same on the other side of the equation. So, we can simply eliminate them to get the net ionic equation: Ag+ (aq) + Cl- (s) -> AgCl (s)Anything that is present in the complete ionic equation, but not in the net ionic equation (the reactants that were also products) are your spectator ions (NO3- and Na+ in this case)And yes, I know that the yield sign should be going both ways, but I don't know how to type that on my iPhone.
How do I identify a spectator ion?
It depends on what kind of reaction you’re looking at.In a redox reaction, a spectator ion is an ion whose oxidation number does not change in the course of the reaction. For example, consider the redox reaction:2Al + 3CuCl2 → 2AlCl3 + 3CuIn this reaction, the chloride ion is a spectator ion since its oxidation number doesn’t change. In order to know that, though, you have to assign oxidation numbers to each of the elements involved in the reaction.You can rewrite the above equation without the chloride ion:2Al + 3Cu^2+ → 2Al^3+ + 3CuIn a double replacement reaction, there are no oxidation number changes. In these situations, spectator ions are ions that do not combine to form insoluble precipitates. Consider the following double replacement reaction:AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)When an ionic compound is dissolved in water, its ions dissociate, so you could write the equation like this:Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq)Now you’ll notice that the sodium (Na+) and nitrate (NO3-) ions are present on both sides of the equation, which means they can be removed from the equation. The sodium and nitrate ions are spectator ions in this case — nothing happens to them from one side of the equation to the other. Here’s the net ionic equation:Ag+(aq) + Cl-(aq) → AgCl(s)
Give the complete ionic equation?
23) Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of lithium sulfide and copper (II) nitrate are mixed. A) Li+ (aq) + SO42-(aq) + Cu+(aq) + NO3-(aq) → CuS(s) + Li+(aq) + NO3-(aq) B) 2 Li+(aq) + S2-(aq) + Cu2+(aq) + 2 NO3-(aq) → CuS(s) + 2 Li+(aq) + 2 NO3-(aq) C) 2 Li+(aq) + S2-(aq) + Cu2+(aq) + 2 NO3-(aq) → Cu2+(aq) + S2-(aq) + 2 LiNO3(s) D) Li+ (aq) + S-(aq) + Cu+(aq) + NO3-(aq) → CuS(s) + LiNO3(aq) E) No reaction occurs. The answer is A BUT, i'm so sad because i don't see why!! Here is what I did: Na2CO3 + HCl → NaCl + H2CO3 (the solid, since no solubility exception is for Carbonate is with Hydrogen) 2Na + CO3^2- + H2 + Cl2 → 2Na + Cl2 + H2CO3 What Cancels: Na, and Cl what it should look like, I think H2 + CO3^2- → H2CO3 (s) What did I do wrong?
Which of these are spectator ions?
The spectator ions are the ones that are the same on each side. In this case that is 2K+ and 2NO3- (the nitrate polyatomic ion - you wrote it incorrectly above). Just like in Algebra, since they are the same on each side they will cancel out on each side (i.e., Reactant and Product side). Thus the Net Ionic Reaction is: Pb2+ (aq) + S2- (aq) --> PbS (s)