Consider the reaction of 30.0 mL of 0.277 M Bal, with 20.0 mL of 0.315 M Na3PO4. Since the precipitate is Ba3(PO4)2, write the net ionic equation for the precipitation reaction. Ba* (aq) + 2 I-(aq) + 6 Na*(aq) + PO4(aq) → Ba3 (PO4) 2 (s) + 6 Na* + 2
Consider the reaction of 30.0 mL of 0.277 M Bal, with 20.0 mL of 0.315 M Na3PO4. Since the precipitate is Ba3(PO4)2, write the net ionic equation for the precipitation reaction. Ba* (aq) + 2 I-(aq) + 6 Na*(aq) + PO4(aq) → Ba3 (PO4) 2 (s) + 6 Na* + 2
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Chapter1: Chemical Foundations
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Consider the reaction of 30.0 mL of 0.277
M Bal, with 20.0 mL of 0.315 M Na3PO4.
Since the precipitate is Ba3(PO4)2, write the
net ionic equation for the precipitation
reaction.
Ba2* (aq) + 2 |-(aq) + 6 Na*(aq) + PO4(aq) →
Вaз (РОa)2(s) +6 Na* + 2
3
3C2
2-
Reset
>
1
2
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6
7
9
(s)
(1)
(g)
(aq)
Na
Ва
NR
• x H2O
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LO
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Transcribed Image Text:Question 21.c of 30
Submit
Consider the reaction of 30.0 mL of 0.277
M Bal, with 20.0 mL of 0.315 M Na3PO4.
Since the precipitate is Ba3(PO4)2, write the
net ionic equation for the precipitation
reaction.
Ba2* (aq) + 2 |-(aq) + 6 Na*(aq) + PO4(aq) →
Вaз (РОa)2(s) +6 Na* + 2
3
3C2
2-
Reset
>
1
2
3
4
6
7
9
(s)
(1)
(g)
(aq)
Na
Ва
NR
• x H2O
Tap here or pull up for additional resources
00
LO
+
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