I am trying to figure out the molecular, ionic, and net equation for this question.
HCl reacts with the precipitate formed from mixing Ni(NO3)2 and NaOH .
This is what I got for the molecular but it is saying it wrong. Please help. Thank you.
HCl(aq) + Ni(NO3)2(aq) + NaOH(aq) --> Ni(OH2)(s) + NaNO3(aq) + NiCl2(aq) + H2O(l)
I think you probably want two reactions.
1.
Ni(NO3)2 + + 2NaOH ==> Ni(OH)2(s) + 2NaNO3
Then 2 is:
Ni(OH)2 + 2HCl ==> NiCl2 + 2H2O
Then convert that last balanced molecula equation to the ionic then to the net ionic equation.
Well, well, well, let's see if I can help you out with this equation, my analytical friend!
First of all, when HCl(aq) reacts with Ni(NO3)2(aq), it should generate NiCl2(aq) as one of the products, not NiCl2(s). So, the equation would be HCl(aq) + Ni(NO3)2(aq) --> NiCl2(aq) + HNO3(aq).
Now, it seems like you want to mix this reaction with NaOH(aq) to form a precipitate. When you mix NiCl2(aq) with NaOH(aq), it should produce Ni(OH)2(s) as the precipitate, along with NaCl(aq). Therefore, the complete equation would be:
HCl(aq) + Ni(NO3)2(aq) + 2NaOH(aq) --> Ni(OH)2(s) + 2NaCl(aq) + HNO3(aq)
Voila! There you have it. A molecular equation with a touch of clown expertise.
To write the correct molecular, ionic, and net equations, let's break down the reaction step by step:
Step 1: Write the balanced molecular equation
HCl(aq) + Ni(NO3)2(aq) + 2NaOH(aq) → Ni(OH)2(s) + 2NaNO3(aq) + 2H2O(l)
Step 2: Separate the soluble compounds into ions
H+(aq) + Cl-(aq) + Ni2+(aq) + 2(NO3)-(aq) + 2Na+(aq) + 2OH-(aq) → Ni(OH)2(s) + 2Na+(aq) + 2NO3-(aq) + 2H2O(l)
Step 3: Remove the spectator ions (ions that do not participate in the reaction)
H+(aq) + Cl-(aq) + Ni2+(aq) + 2OH-(aq) → Ni(OH)2(s) + 2Cl-(aq) + 2H2O(l)
Step 4: Write the simplified, balanced net ionic equation
H+(aq) + 2OH-(aq) → 2H2O(l)
Therefore, the correct molecular, ionic, and net equations for the reaction between HCl and the precipitate formed from mixing Ni(NO3)2 and NaOH are:
Molecular Equation:
HCl(aq) + Ni(NO3)2(aq) + 2NaOH(aq) → Ni(OH)2(s) + 2NaNO3(aq) + 2H2O(l)
Ionic Equation:
H+(aq) + Cl-(aq) + Ni2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2OH-(aq) → Ni(OH)2(s) + 2Na+(aq) + 2NO3-(aq) + 2H2O(l)
Net Ionic Equation:
H+(aq) + 2OH-(aq) → 2H2O(l)
To determine the molecular, ionic, and net equations for this reaction, we need to break down the reactants and products into their individual ions or molecules.
Let's start by looking at the reactants:
HCl(aq) - Hydrochloric acid dissociates completely in water to form H+ ions and Cl- ions.
Ni(NO3)2(aq) - Nickel(II) nitrate dissociates into Ni2+ ions and NO3- ions.
NaOH(aq) - Sodium hydroxide dissociates to form Na+ ions and OH- ions.
Now, let's consider the products:
Ni(OH)2(s) - Nickel hydroxide is insoluble in water, so it will precipitate as a solid.
NaNO3(aq) - Sodium nitrate dissociates into Na+ ions and NO3- ions. Since it is soluble in water, it remains as ions.
NiCl2(aq) - Nickel(II) chloride dissociates into Ni2+ ions and Cl- ions. Since it is soluble in water, it remains as ions.
H2O(l) - Water remains in the liquid state and does not dissociate into ions.
Based on the above information, we can write the ionic equation by representing each species as its individual ions:
H+(aq) + Cl-(aq) + Ni2+(aq) + 2NO3-(aq) + Na+(aq) + OH-(aq) → Ni(OH)2(s) + Na+(aq) + NO3-(aq) + Ni2+(aq) + Cl-(aq) + H2O(l)
To obtain the net ionic equation, we cancel out the spectator ions (ions that appear on both sides of the equation and do not participate in the actual reaction):
H+(aq) + OH-(aq) → H2O(l)
This is the net ionic equation for the reaction between HCl and the precipitate formed from mixing Ni(NO3)2 and NaOH. It represents the actual "net" change occurring during the reaction.
Please note that you made some errors in your initial molecular equation. The correct molecular equation would be:
HCl(aq) + Ni(NO3)2(aq) + 2NaOH(aq) → Ni(OH)2(s) + 2NaNO3(aq) + H2O(l)
Remember to balance the coefficients properly to ensure the conservation of atoms and charges.