balance the folliwing equation by algebrac method
Answers
Answer:
Explanation:
The strategy for balancing chemical equations algebraically is as follows:
Write a different letter coefficient in front of each compound in the equation
Write algebraic expressions or rules for each element that equate its atoms on the LHS and RHS
Substitute and simplify to obtain a rule that equates only two letter coefficients that you can solve
Substitute the values into the other rules to obtain the balancing coefficients
That makes absolutely no sense I’m sure without an example, so here’s an equation to balance using this strategy:
_KMnO4 + _HCl → _MnCl2 + _KCl + _Cl2 + _H2O
First thing we do is give each compound a letter coefficient:
aKMnO4 + bHCl → cMnCl2 + dKCl + eCl2 + fH2O
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.
K: a = d
Mn: a = c
O: 4a = f
H: b = 2f
Cl: b = 2c + d + 2e
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. On the reactant side, we would have a total of b chlorine atoms. On the product side, MnCl2 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.
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:
b = 2a + a + 2e
b = 3a + 2e
We can also get rid of b using the rule for H:
2f = 3a + 2e
And finally, get rid of f using the rule for O:
2(4a) = 3a + 2e
8a = 3a + 2e
5a = 2e, hence a = 2 and e = 5
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:
b = 2c + d + 2e
b = 3a + 2e
b = 3 x 2 + 2 x 5
∴ b = 16
4a = f
∴ f = 8
2KMnO4 + 16HCl → 2MnCl2 + 2KCl + 5Cl2 + 8H2O
In the unlikely event this showed up on an exam paper as a balancing question, you would ordinarily use the redox balancing method but I think you’d agree that this method is faster.
Note that when you use the algebraic method, you may end up performing different substitutions in order to eliminate the unknowns. This is perfectly fine, there is no right or wrong approach.