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Chemical Equations - h1015v1_63
Chemical Equations - h1015v1_65

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







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