Asoom Posts: 52, Reputation: 1 Junior Member #1 Apr 20, 2009, 12:58 PM
Write the balance equation
I tried to solve my homework , but I face some problem Im not sure about it and another I can't solve it..

Im not sure about the solve , could you check if it is right or wrong

1- (Salt + dil. HCl) with Carbonate

CO3^2- + 2HCl -----> CO2 + H2O+ 2Cl^-

2- (Solution + HgCl2) with Crbonate

CO3^2- + HgCl2 -----> HgCO3 + 2Cl^-

3- (Solution +MgSO4) with Crbonate

CO3^2- + MgSO4 -----> MgCO3 + SO4^2-

4- (Salt + dil. HCl) with Bicarbonate

HCO3^2- + HCl -----> CO2 + H2O+ Cl^-

5- (Solution +HgCl2) with Bicarbonate

HCO3^2- + HgCl2 -----> HgCO3 + HCl

6- (Solution +MgSO4) with Bicarbonate

HCO3^2- + MgSO4-----> MgCO3+ HSO4

7- (Salt + dil. HCl) with Nitrite

NO2^- + HCl -----> HNO3 + Cl

8- (Solution +AgNO3) with Nitrite

NO2^- + AgNO3 -----> AgNO2 + NO3

I can't solve it

1- (Solution + acidified KMnO4) with Nitrite
2- (Salt + dil. HCl) with thiosulphate
3- (Solution +AgNO3) with thiosulphate
4- (Solution + acidified KMnO4) with thiosulphate
 Perito Posts: 3,139, Reputation: 150 Ultra Member #2 Apr 20, 2009, 01:37 PM
1- (Salt + dil. HCl) with Carbonate

$CO_3^{-2} + 2HCl -----> CO_2 + H_2O+ 2Cl^-$
Correct.

2- (Solution + HgCl2) with Crbonate

$CO_3^{-2} + HgCl_2 -----> HgCO_3 + 2Cl^-$
Correct

3- (Solution +MgSO4) with Crbonate

$CO_3^{-2} + MgSO_4 -----> MgCO_3 + SO_4^{2-}$
Correct

4- (Salt + dil. HCl) with Bicarbonate

$HCO_3^- + HCl -----> CO_2 + H_2O+ Cl^-$
Almost correct It's HCO3^- not HCO3^2-. There's one negative charge, not 2.

5- (Solution +HgCl2) with Bicarbonate

$HCO_3^- + HgCl_2 -----> HgCO_3 + HCl + Cl^-$
You were missing one chlorine and one negative charge.

6- (Solution +MgSO4) with Bicarbonate

$HCO_3^- + MgSO_4-----> MgCO_3+ HSO_4^-$
Again, bicarbonate has only one charge. Bisulfate also has a -1 charge.

7- (Salt + dil. HCl) with Nitrite

$NO_2^- + HCl -----> HNO_2 + Cl^-$
You must also make sure the charge is balanced.

8- (Solution +AgNO3) with Nitrite

$NO2^- + AgNO_3 -----> AgNO_2 + NO_3^-$
Again, balance charge.

1- (Solution + acidified KMnO4) with Nitrite
This is an oxidation-reduction reaction. Nitrite is oxidized, Permanganate is reduced.

2- (Salt + dil. HCl) with thiosulphate
Thiosulfate decomposes in acid solution to form SO2 and sulfur.

3- (Solution +AgNO3) with thiosulphate
Forms silver thiosulfate.

4- (Solution + acidified KMnO4) with thiosulphate
thiosulfate is a reducing agent, permanganate is an oxidizing agent. The thiosulfate will be oxidized, the permanganate will be reduced.

Try to figure out these last 4 with the hints I've given you.
 Asoom Posts: 52, Reputation: 1 Junior Member #3 Apr 20, 2009, 10:18 PM

Ok , I will try after my classes finish today

Thanks a lot
 Asoom Posts: 52, Reputation: 1 Junior Member #4 Apr 21, 2009, 12:32 PM

1- NO2^- + KMnO4 ------> KNO2 + MnO4

2- S2O3^2- + 2HCl ------> SO2 + S + Cl^- + H2O

3- S2O3^2- + 2AgNO3 ------> Ag2S +2NO3

the last one I can't solve it

(salt + conc H2SO4)

4I^- + H2SO4 -------> I2 + 2HI + SO4
could you check also this equation
 Perito Posts: 3,139, Reputation: 150 Ultra Member #5 Apr 21, 2009, 01:11 PM
$NO_2^- + KMnO_4 -\rightleftharpoons KNO_3 + MnO_2$
NO2(-) is oxidized to NO3(-). MnO4- is reduced either to MnO2 or to Mn(2+) or occasionally to MnO4(-2). I haven't balanced this and you might need water, acid, or base to do so.
Permanganate - Wikipedia, the free encyclopedia

Balancing by half reactions
$MnO_4^- + 2H^+ + e^- \rightarrow MnO_2 + 2H_2O$

$NO_2^- + H_2O \rightarrow NO_3^- + 2H^+ + 2e-$

Multiply the first equation by 2 and add to the second equation

$2MnO_4^- + 2H^+ + NO_2^- \rightarrow 2MnO_2 + 3H_2O + NO_3^-$

That should be balanced.

$S_2O_3^{2-}+ + 2HCl ------> SO_2 + S + Cl^- + H_2O$
looks right.

$S_2O_3^{2-} + 2AgNO_3 ------> Ag_2S_2O_3+2NO_3^-$
I think you just form silver thiosulfate. I don't think the thiosulfate is strong enough to reduce the silver. If it does, it will form metallic silver.

(salt + conc H2SO4)

$2I^- + H_2SO_4 \rightleftharpoons 2HI + SO_4 ^{-2}$
There's no oxidation. Sulfuric acid is not an oxidizer so you can't oxidize I- to I2.
In fact, you'll have an equilibrium with "bisulfate" and "sulfate":

$2I^- + H_2SO_4\, \rightleftharpoons\, I^- + HI + HSO_4 ^-\, \rightleftharpoons\, 2HI + SO_4^{-2}$
 Asoom Posts: 52, Reputation: 1 Junior Member #6 Apr 21, 2009, 10:03 PM

Thanks a lot , your explanation was very strong and clear..

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