Question

Balance the following chemical equations using the algebraic method. Make sure you show all your work and explain each the step.

(1) C2H2 + O2, reaction arrow, CO2 + H2O

Answer 1

$C_2H_2 + O_2 \rightarrow CO_2 + H_2O$

Firstly, we give each compound a coefficient

$aC_2H_2 + bO_2 \rightarrow cCO_2 + dH_2O$

Next, applying the Conservation of Mass law, which tells us that the total number of atoms of each element must be the same on both sides, we then write;

$\begin{aligned} C\ (Carbon) &: 2a = c\\ H\ (Hydrogen) &: 2a = 2d\\ O\ (Oxygen) &: 2b = 2c + d \end{aligned}$

Note that the amount of reactant occupy the first part of the ratio and the amount of the product the second part

$\begin{aligned} \therefore \textsf{we have}\\ 2a &= c \ ---(i)\\ 2a &= 2d \ ---(ii)\\ 2b &= 2c + d ---(iii)\\ \end{aligned}$

As there are no free terms in this set of equations, it has a trivial solution. But in the case of chemical equations we have one major information -> All coefficients must be integer and they must be the smallest ones. To find them we can assume one of the coefficients to be 1. Therefore let's assume coefficient a to be 1

$\\ \textsf{From equation }(i),\\ 2a = c\\ 2(1) = c\\ \therefore c= 2$

$\\ \textsf{From equation }(ii),\\ 2a = 2d\\ \therefore a = d\\ \textsf{but }a= 1, \ \therefore d= 1$

$\textsf{From equation }(iii),\\ 2b = 2c + d\\ \textsf{but }c = 2, \textsf{and }d = 1\\ \therefore 2b = 2(2) + 1\\ 2b = 5\\ b = \frac{5}{2}$

$\therefore a = 1, \ b = \frac{5}{2},\ c = 2,\ d = 1$

But we know that the coefficients have to be integers, so we multiply all values by 2

$\therefore$ the final value of each coefficients become;

$a = 2, \ b = 5,\ c = 4,\ d = 2$

$\textsf{The balanced chemical reaction is; }\\ 2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O$