The choice of which coefficient to set to 1 is completely arbitrary and may be taken to be the one that is most convenient for solving the problem. Thus, we may take one of the coefficients to be 1, arbitrarily, which then leaves only three unknowns and three conditions to determine them. Thus, one of the coefficients is arbitrary, and all the others can be expressed relative to it. The fourth condition comes from recognizing that chemical equations specify relative amounts of reactants and products. This constitutes a set of only three equations for the four unknowns. From carbon balance, we have the condition: In this case, there are are four unknowns, x, y, z and w. Then we write down the balance conditions for each element in terms of the unknowns. To balance a reaction algebraically, we start by putting unknown coefficients in front of each molecular species in the equation: However, we will use it to illustrate another approach – the algebraic approach. This equation could be balanced by inspection. Balancing by inspection in this way is quick and useful when you can see the solution easily. Now there are equal quantities of all elements on both sides of the reaction, and mass conservation is satisfied. The balancing of oxygen and hydrogen can be handled together by making the coefficient of H 2O equal to 2. Similarly, there are 4 hydrogens on the left and only 2 on the right, so hydrogen is not balanced. There are 3 oxygens on the left but only 2 on the right, so oxygen is not balanced. We notice that there are 2 nitrogens on both the left and right sides, so nitrogen is already balanced.
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