My teacher hasn't really talked about types of reactions; he's just given us a list of net ionic equations involving water and different elements including Lithium, Sodium, Calcium, Magnesium, Lanthanum, and Potassium to memorize. However, his test questions don't involve the reactants/net ionic equations he's given us, so when I see "Show a net ionic equation for the reaction between Water and Strontium metal" I don't know what to do.
Elements in Group IIA of the periodic table can replace one of the hydrogen atoms in H2O to produce a hydroxide and H2 gas. Example:
Ca(s) + H2O ---> Ca(OH)2 + H2
Ca(s) + 2H2O ---> Ca(OH)2 + H2 (balanced)
Other elements in the same column as Ca react in a similar way.
Elements in Group IA of the periodic table can replace one of the hydrogen atoms in H2O, also, to produce a hydroxide and H2 gas. Example:
Li + H2O ---> LiOH + H2
2Li + H2O ---> 2LiOH + H2 (balanced)
Other metals in the same column as Li react in a similar way.
La reacts in a similar way but the products are La(OH)3 and H2. Try writing and balancing the chemical equation.
Has your teacher given you the molecular equation. For Na, that is
Na + H2O ==> NaOH + H2 and balanced it will be
2Na + 2H2O ==> 2NaOH + H2
Now what you want to do is to show the form as follows. To simplify things, I will not worry about the balancing coefficients.
2Na(s) + 2H2O(l) ==> 2NaOH(aq) + H2(g)
s=solid
l=liquid
aq = aqueous
g=gas
Now you want to take the molecular equation and make that into an ionic one.
solids don't change.
liquids don't change
aq (separate into ions).
gases don't change.
Na(s) + H2O(l) ==> Na^+(aq) + OH^-(aq) + H2(g)
Sr metal, Ca metal, K metal, any of the metals that will react with water will do so to produce the corresponding base (the OH part) plus H2(g).
All of the ion part will be analogous. The final step in making a NET ionic equation is to subtract (cancel) those items that are common to both sides. In the above equation, the final ionic equation IS the net ionic equation because nothing cancels.
I will do the Pb(NO3)2 and LiCl, again but I will use the balanced equation.
The molecular equation is
Pb(NO3)2 + 2LiCl ==> PbCl2 + 2LiNO3
You had aqueous solution so I will add that now.
Pb(NO3)2(aq) + 2LiCl(aq)==>PbCl2(s) + 2LiNO3(aq)
Now change these to ions.
Pb(NO3)2 is soluble; therefore, it becomes the ions, LiCl ditto. PbCl2 is insoluble and stays as is. LiCl is soluble and changes to ions.
Pb^+2(aq) + 2NO3^-(aq) + 2Li^+(aq) + 2Cl^-(aq) ==> PbCl2(s) + 2Li^+(aq) + 2NO3^-(aq).
That last is the molecular equation changed to the ionic equation. Now to make the net ionic equation, we cancel those items common to both sides. In this case there is no Pb^+2 on the right so we keep it on the left. Notice there are 2NO3^- on the left and right; therefore, cancel them. They will not show up in the final equation. There are 2Li^+ on both sides so cancel them. They will not show up either. What's left?
Pb^+2(aq) + 2Cl^-(aq) ==> PbCl2(s)
This is the net ionic equation. Sorry it took me so long but this isn't easy to type.
I appreciate your patience! Thank you.
thanks!
When faced with a question about a specific reaction, such as the one between water and strontium metal, you can use some principles of reactivity to help you derive the net ionic equation.
Here's a step-by-step approach to determine the net ionic equation between water and strontium metal:
1. Understand the Reactants: In this case, you are given water (H2O) and strontium metal (Sr). Water consists of hydrogen (H) and oxygen (O), while strontium metal exists as strontium ions (Sr2+) in water.
2. Identify the Possible Reactivity: Look at the reactivity series of metals, which ranks them in terms of their reactivity with water. Strontium (Sr) is a highly reactive metal, so it is likely to react with water.
3. Determine the Products: When a reactive metal reacts with water, it displaces hydrogen gas (H2) and forms a metal hydroxide. In this case, the product is strontium hydroxide (Sr(OH)2) and hydrogen gas (H2).
4. Write the Balanced Equation: Start by writing the balanced chemical equation for the reaction between water and strontium metal:
Sr + 2H2O -> Sr(OH)2 + H2
5. Identify the Spectator Ions: The spectator ions are the ions that do not participate in the reaction and remain the same on both sides of the equation. In this case, the spectator ions are the hydroxide ions (OH-) since they are present on both sides.
6. Write the Net Ionic Equation: Remove the spectator ions from the equation, and you will be left with the net ionic equation:
Sr + 2H2O -> Sr2+ + 2OH- + H2
The final net ionic equation for the reaction between water and strontium metal is Sr + 2H2O -> Sr2+ + 2OH- + H2.