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BALANCING CHEMICAL EQUATIONS USING THE ALGEBRAIC METHODLet's try an algebraic method forH3BO3→ H4B6O11 + H2O It can be rather easily balanced by inspection, but let's try a more systematic approach.What does 'balanced' mean? It means that for every element, there is the same number of atoms on both sides of the reaction equation. Our reaction has three coefficients a, b and c:aH3BO3→bH4B6O11 + cH2O 'Balanced' means that there is exactly the same number of atoms of each element on both sides of theequation. Using coefficients a, b and c we can tell that we have 1×a atoms of boron on the left (one atom per each H3BO3 molecule), and 6×b + 0×c on the right (6 atoms of boron per each H4B6O11 molecule and no boron in water). This gives us the following equation:1×a = 6×b + 0×cWe can write similar equations for all elements - hydrogen:3×a = 4×b + 2×cand oxygen:3×a = 11×b + c As there are no free terms in this set of equations, it has a trivial solution (a = b = c = 0) which we are not interested in. We have three equations, and three unknowns - nothing particularly difficult to solve. Quite often you will end with many more equations and many more unknowns. Such equation sets is not a thing that you may want to solve manually, although when balancing chemical equations in most cases it can be done relatively easy, as most equations don't contain all unknowns. In this case we have a very simple equation a = 6×b that we can use to substitute 6×b for a in the second and third equation to get:18×b = 4×b + 2×c18×b = 11×b + cAfter some rearranging:7×b = c7×b = cBoth equations are identical. In algebra it usually means that the set of equations doesn't have a unique solution, but in the case of chemical equations we have one additional information - all coefficients must
be an integer and they must be the smallest ones. To find them we can assume one of the coefficients to be 1:b = 1If soa = 6c = 7and indeed6H3BO3→ H4B6O11 + 7H2O is the balanced reaction equation.This first example doesn't look convincing - why do we have to solve a set of equations when the reaction equation can be easily balanced by other means? Good point - but what if the reaction can be not easily balanced?P2I4 + P4 + H2O → PH4I + H3PO4Try for a moment. Looks easy but soon gets surprisingly hard and the coefficients become pretty high, which makes you wonder if you have not made some mistake (*see some balancing hints at the bottom of the page). What about the general, algebraic method?We need five coefficients, and there are only four equations (one for each element present) - but it shouldn't bother us, as we know that we have additional information that works as an additional equation.aP2I4 + bP4 + cH2O →dPH4I +eH3PO4Setting up equations:P: 2×a + 4×b = d + eI: 4×a = d H: 2×c = 4×d + 3×eO: c = 4×eBalances for iodine and oxygen make this set look much easier than expected. c = 4×e and d = 4×a are substitutions that we are about to use to reduce number of unknowns:1st equation: 2×a + 4×b = 4×a + e3rd equation: 8×e = 4×d + 3×e