1996 United States House of Representatives elections in Kansas|
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|      |  | Majority party | Minority party |   |  |  |  |   | Party | Republican | Democratic |   | Last election | 4 | 0 |   | Seats won | 4 | 0 |   | Seat change |   |   |   | Popular vote | 591,146 | 424,984 |   | Percentage | 56.36% | 40.52% |   | Swing |  7.15% |  4.03% |  | 
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  District results County results     | Republican    40–50%   50–60%   60–70%   70–80%   80–90%   >90% | Democratic    50–60%   60–70%    |  | 
  The 1996 United States House of Representatives elections in Kansas were held on November 5, 1996, to determine who will represent the state of Kansas in the United States House of Representatives. Kansas has four seats in the House, apportioned according to the 1990 United States census. Representatives are elected for two-year terms. 
 Overview
    | United States House of Representatives elections in Kansas, 1996[1] | 
  | Party | Votes | Percentage | Seats | +/– | 
  |  | Republican | 591,146 | 56.36% | 4 | - | 
  |  | Democratic | 424,984 | 40.52% | 0 | - | 
  |  | Libertarian | 23,253 | 2.22% | 0 | - | 
  |  | Reform | 9,495 | 0.91% | 0 | - | 
   | Totals | 1,048,878 | 100.00% | 4 | - | 
 
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