1. net ionic equation for the reaction of tin (II) and solid mercury chloride to yield tin (IV), liquid mercury and chloride ion
2. net ionic equation for the reaction between solid SbOOH and H2S
I am not too sure about my equations, so I was wondering if someone could check with me, and if something was wrong, can you teach me how to do it.
1. Hg (2+) + Sn (2+) ---> Hg (l) + Sn (4+)
2. 2SbOOH + 2H2S ----> Sb2OS2 + 3 H2O
for some reason, they don't seem write to me.
The two chemical equations are balanced for electrical charge and number of atoms. One additional requirement is that they actually occur. That is a little more difficult to ascertain.
but are they correct in essence of net ionic equation?
The first one probably is ok. It often occurs when we don't want it to occur in the reduction step of Fe^+3 with SnCl2. If we add too much SnCl2 in excess, then we get the black Hg ppt of solid Hg; then that replicate sample must be discarded. The other one, I agree with GK, doesn't appear to make sense. I would think it more likely that Sb2S3 or Sb2S5 would ppt.
To write the net ionic equations, we need to consider the charges of the involved ions. Let's go through each reaction step by step:
1. Reaction of tin (II) and solid mercury chloride:
First, we need to identify the chemical formulas of the reactants:
Tin (II) ion: Sn^2+
Mercury chloride: HgCl2
The next step is to determine the products of the reaction. Tin (IV), liquid mercury, and chloride ions are formed:
Tin (IV) ion: Sn^4+
Liquid mercury: Hg(l)
Chloride ion: Cl^-
To balance the equation, we need to ensure that the charges are equal on both sides. In this case, we have two chloride ions on the reactant side, so we need two chloride ions on the product side:
HgCl2 + Sn^2+ → Hg(l) + Sn^4+ + 2Cl^-
Now, let's write the balanced net ionic equation. To do this, we only include the species that participate in the reaction:
Hg^2+ + Sn^2+ → Hg(l) + Sn^4+
2. Reaction between solid SbOOH and H2S:
For this reaction, we need to identify the chemical formulas of the reactants:
Solid SbOOH: SbOOH
Hydrogen sulfide: H2S
The products are Sb2OS2 and water:
Sb2OS2
Water: H2O
To balance the equation, we need to ensure that the number of each type of atom is the same on both sides. Let's examine the reaction:
SbOOH + 2H2S → Sb2OS2 + 3H2O
Now, let's write the balanced net ionic equation. This equation only includes the species that actively participate in the reaction:
SbOOH + 2H2S → Sb2OS2 + 3H2O
Both of your net ionic equations were incorrect, but don't worry, writing net ionic equations can be tricky. By following the steps above, you should be able to write accurate net ionic equations.