Having now found two coefficients, substituting into the rules for Cl and O and using the fact that a = c = d = 2 solves for b and f:ĢKMnO 4 + 16HCl → 2MnCl 2 + 2KCl + 5Cl 2 + 8H 2O We can also get rid of b using the rule for H:Īnd finally, get rid of f using the rule for O: There are too many unknowns here, but we can substitute the rules for K and Mn into the rule for Cl to get rid of c and d: On the product side, MnCl 2 contains two chlorine atoms, so if its coefficient is c, it must contain 2c chlorine atoms, whilst KCl contains d chlorine atoms and so on, adding up the total number of chlorine atoms on the RHS. On the reactant side, we would have a total of b chlorine atoms. To explain the logic behind this using Cl as an example, we know that the number of chlorine atoms must be the same on both sides of the equation. Next, applying the Conservation of Mass, which tell us that the total number of atoms of each element must be the same on both sides, write algebraic rules for each element. _KMnO 4 + _HCl → _MnCl 2 + _KCl + _Cl 2 + _H 2Oįirst thing we do is give each compound a letter coefficient:ĪKMnO 4 + bHCl → cMnCl 2 + dKCl + eCl 2 + fH 2O That makes absolutely no sense I’m sure without an example, so here’s an equation to balance using this strategy:
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