2 years ago

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Chemistry

1 answer:

Sergeeva-Olga [200]2 years ago

5 0

Answer: I have no way of answering this ,let alone give you the right answer

Explanation: You din't even tell us what the Q is

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"solid potassium iodide decomposes into iodine gas and solid potassium. Write a a balanced chemical equation for this reaction"

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$KI (s) ----\ \textgreater \ K(s) + I_2 (g)$

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$2KI (s) ----\ \textgreater \ 2K(s) + I_2 (g)$

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When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o

Feliz [49]

Answer:

The net ionic equation for the given reaction :

$3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)$

Explanation:

$3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)$ ...[1]

$MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)$ ..[2]

$Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)$ ...[3]

$NaI(aq)\rightarrow Na^+(aq)+I^-(aq)$

Replacing $MnI_2(aq)$ , NaI and $Na_3PO_4(aq)$ in [1] by usig [2] [3] and [4]

$3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)$

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

$3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)$

8 0

2 years ago