1somebody plz answer this
balance the reactions:
(i) SO2 + Na2CrO4 + H2SO4 → Na2SO4 + Cr2(SO4)3 +H2O
(ii) C2H5OH + I2 + OH- → CHI3 + HCO2- + H2O + I-
I am facing the same problem in both the reactions
for eg taking (i) reaction
step I
S4+ → S6+ + 2e-
6e- + 2Cr6+ → (Cr3+)2
step II
combining the reactions in step I we get
3S4+ + 2Cr6+ → 3S6+ + (Cr3+)2
now when we write the S6+ on product side with its compounds in original reaction we have to distribute it in two compounds that are
Na2SO4 & Cr2(SO4)3 ........how can i know that how many moles of S6+ go to Na2SO4 and how many to other ??
Same is the problem with second reaction in which we have to distribute
C2+ in CHI3 & HCO2- and also distribute
I- in CHI3 & I-
plz explain how do we carry out such distributions and if any other way of balancing these reactions
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5 Answers
13571.
Reactant is 2Na2CrO4, So on product side it should be 2Na2SO4 (balancing of Na)
1sir why are we looking to balance only Na
although your answer is right
if we use 2Na2SO4 then S will be unbalanced
23for 1st , after writing step 2 ,
u shud write the respective compounds of the species that are in step 2 only ,
for 3S6+ & 2Cr3+ , as they are in the same compound, u shud write that compound only ,bcz if u would hav written Na2SO4 , then while balancing Na atoms, u will be adding Na2CrO4 , but since Cr is already balanced , u shud not do that .
so
3SO2 + 2Na2CrO4 → Cr2(SO4)3
now balance Na atoms
3SO2 + 2Na2CrO4 → 2Na2SO4 + Cr2(SO4)3
now using H2SO4 balance SO4 grp of 2Na2SO4 only bcz u already balanced sulphur atoms of Cr2(SO4)3 while balancing with e-
so
3SO2 + 2Na2CrO4 + 2 H2SO4 → 2Na2SO4 + Cr2(SO4)3
balance H now and check if O atoms are also getting balanced simultaneously
3SO2 + 2Na2CrO4 + 2 H2SO4 → 2Na2SO4 + Cr2(SO4)3 + 2H2O
232.
C2- → C2+ + 4e-
I2 + 2e- → 2 I-
so C2- + 2 I2 → C2+ + 4I-
multiply by 2
2C2- + 4 I2 → 2 C2+ + 8I-
i.e
2C2- + 4 I2 → (C2+ + 3I- ) + C2+ + 5 I-
write CHI3 for (C2+ + 3I- ) , HCO2- for C2+,
so
C2H5OH + 4 I2 → CHI3 + HCO2- + 5 I-
balance charge using OH-
C2H5OH + 4 I2 + 6OH- → CHI3 + HCO2- + 5 I-
now balance O atoms using H2O
C2H5OH + 4 I2 + 6OH- → CHI3 + HCO2- + 5 I- + 5 H2O