CHEMICAL EQUATIONSDOE-HDBK-1015/1-93Fundamentals of ChemistryCH-01Rev. 0Page 42Example 1:FeS + O Fe O + SO2 2 2 3 2Solution 1:Starting with Fe O (see guideline b), write 2FeS + O Fe O + SO , which2 3 2 2 2 3 2balances the Fe on each side of the equation. Now there are 4 S atoms on the left side so balance the S atoms by writing2FeS + O Fe O + 4SO . 2 2 2 3 2Everything is balanced except the O. There are now 2 O atoms on the left and 11 Oatoms on the right. To get 11 O atoms on the left write2FeS + 5.5O Fe O + 4SO . 2 2 2 3 2This makes both sides of the equation balanced except the coefficients must be wholenumbers (guideline f). To meet guideline f, multiply both sides by two which will bringthe left side to a whole number of O molecules. 2Thus, the solution is 4FeS + 11O 2Fe O + 8SO .2 2 2 3 2Example 2:NH + CuO H O + N + Cu3 2 2Solution 2:Start with NH since there are two N atoms on the right of the equation. To balance the3N atoms write 2NH + CuO H O + N + Cu (guideline b).3 2 2Since the H appears in only the NH and H O and the NH has been balanced, the H O3 232will be balanced. Write 2NH + CuO 3H O + N + Cu (guideline e).322Oxygen appears only in CuO and in H O, and since the H O has been already been22balanced write 2NH + 3CuO 3H O + N + Cu (guideline e).322That leaves the Cu to be balanced. Thus, the solution is2NH + 3CuO 3H O + N + 3Cu.322
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