Asked by Donna
For a reaction between sodium phosphate and strontium nitrate write out the following:
a) The balanced molecular equation
b) The ionic equation
c) The net ionic equation
a) I think I have this one:
2 Na3PO4 (aq) + 3 Sr(NO3)2 (aq) --> 6 NaNO3 (aq) + Sr(PO4)2 (s)
b) Should I just take apart the unbalanced equation?
3Na^+(aq) + PO4^3-(aq) + 2NO3^-(aq) --> Na+(aq) + NO3^-(aq) + Sr3(PO4)2(s)
c) If the above is the ionic equation then I have no idea how to make the net ionic equation. I don't see anything canceling out.
a) The balanced molecular equation
b) The ionic equation
c) The net ionic equation
a) I think I have this one:
2 Na3PO4 (aq) + 3 Sr(NO3)2 (aq) --> 6 NaNO3 (aq) + Sr(PO4)2 (s)
b) Should I just take apart the unbalanced equation?
3Na^+(aq) + PO4^3-(aq) + 2NO3^-(aq) --> Na+(aq) + NO3^-(aq) + Sr3(PO4)2(s)
c) If the above is the ionic equation then I have no idea how to make the net ionic equation. I don't see anything canceling out.
Answers
Answered by
Anonymous
for part a, i think you forgot to put Sr3(PO4)2 (since you used that for part b). So, part a's balanced equation should be: 2 Na3PO4 (aq) + 3 Sr(NO3)2 (aq) <--> 6 NaNO3 (aq) + Sr3(PO4)2 (s)
Answered by
Donna
Oh, sorry, I meant Sr3(PO4)2. That's what I wrote down but I didn't type it properly.
Answered by
DrBob222
The Na^+ cancels on both sides and the NO3^- cancels on both sides (you omitted the 6 coefficient for the rigght side for both Na+ and NO3-).
So the net ionic equation is
PO4^-3(aq) + Sr^+2(aq) ==> Sr3(PO4)2(s)
So the net ionic equation is
PO4^-3(aq) + Sr^+2(aq) ==> Sr3(PO4)2(s)
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