# Balancing the equation for the preparation of dichromate from chromite ore

$$\ce{FeCr_2O4 + Na2CO3 + O2 -> Na2CrO4 + Fe2O3 + CO2}$$

I have tried balancing it by writing the reduction and oxidation half equations viz. :

$$\ce{2Fe^{+II}Cr_2^{+III}O4 -> 4Na2Cr^{+VI}O4 + Fe2^{+III}O3 + 14e^-}$$

And:

$$\ce{3O2^{0} + 12 e^- -> 2Fe2O3^{-II} }$$

Then I multiplied the first equation by 6 and second by 7 and added them and though the electrons got cancelled, it turned out to be really hard to make any progress.

Could someone please provide the technique for balancing such an equation?

• $\ce{4FeCr_2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$ This is a simple balance. – Soumik Das Nov 20 '18 at 18:34
• I have the answer @soumikdas. I wish to know the method to arrive at it. – Archer Nov 20 '18 at 18:35
• Balancing is a case of trial and error most of the time and just spotting what works and thinking logically about the equation – H.Linkhorn Nov 20 '18 at 21:12

As of someone mentioned here, balancing most chemical equations is a a case of trial and error. Your equation is easy to balance by this way than go to extremes.

1) A relatively complicated formula within this equation is $$\ce{FeCr2O4}$$ on the reactants' side (LHS) of the equation. Since $$\ce{Fe}$$ atom in it is converted to the resultant product, $$\ce{Fe2O3}$$ and no other $$\ce{Fe}$$ bearing products on products' side (RHS), you should start with putting coefficient $$2$$ in front of $$\ce{FeCr2O4}$$ to balance $$\ce{Fe}$$ in LHS.

2) It would make four chromium atoms in LHS. Thus, you need to balance that by putting coefficient $$4$$ in front of $$\ce{Na2CrO4}$$ on RHS because it is the only $$\ce{Cr}$$ bearing molecule among products.

3) That action would make eight sodium atoms in RHS. Thus, you need to balance that extra $$\ce{Na}$$ by putting a coefficient $$4$$ in front of $$\ce{Na2CO3}$$ on LHS because it is the only $$\ce{Na}$$ bearing molecule among reactants.

3) Now, you would find four carbon atoms in RHS. To balance that extra $$\ce{C}$$ you may need to put a coefficient $$4$$ in front of $$\ce{CO2}$$ in LHS because it is the only $$\ce{C}$$ bearing molecule among products.

4) Now you have balance all atoms except for oxygen atoms. Thus, count the number of oxygen atoms in RHS and LHS, separately. If you count correctly there are 27 $$\ce{O}$$ atoms in RHS and only 20 of them in LHS. To balance that extra 7 $$\ce{O}$$ atoms, you should put a coefficient $$\frac{7}{2}$$ in front of $$\ce{O2}$$ in LHS. This would complete the balancing your equation: $$\ce{2FeCr2O4 + 4Na2CO3 + 7/2 O2 -> 4Na2CrO4 + Fe2O3 + 4CO2}$$

In addition, some people do not like to keep fractions as coefficients in their balance equations. So, you may multify whole equation by 2 to get integers as a coefficients: $$\ce{4FeCr2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$$