含有n克HNO3的稀溶液跟m克Fe恰好反应,铁全部溶解,生成NO,已知有n/4克HNO3被还原,则n:m不可能是A 3:1B 3:2C 4:1D 9:2用极端假设法来写1.若铁被完全氧化成Fe3+ 则根据得失电子守恒有:(m/56)*3=(n/4/63)*3 得n
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![含有n克HNO3的稀溶液跟m克Fe恰好反应,铁全部溶解,生成NO,已知有n/4克HNO3被还原,则n:m不可能是A 3:1B 3:2C 4:1D 9:2用极端假设法来写1.若铁被完全氧化成Fe3+ 则根据得失电子守恒有:(m/56)*3=(n/4/63)*3 得n](/uploads/image/z/5001634-10-4.jpg?t=%E5%90%AB%E6%9C%89n%E5%85%8BHNO3%E7%9A%84%E7%A8%80%E6%BA%B6%E6%B6%B2%E8%B7%9Fm%E5%85%8BFe%E6%81%B0%E5%A5%BD%E5%8F%8D%E5%BA%94%2C%E9%93%81%E5%85%A8%E9%83%A8%E6%BA%B6%E8%A7%A3%2C%E7%94%9F%E6%88%90NO%2C%E5%B7%B2%E7%9F%A5%E6%9C%89n%2F4%E5%85%8BHNO3%E8%A2%AB%E8%BF%98%E5%8E%9F%2C%E5%88%99n%3Am%E4%B8%8D%E5%8F%AF%E8%83%BD%E6%98%AFA+3%3A1B+3%3A2C+4%3A1D+9%3A2%E7%94%A8%E6%9E%81%E7%AB%AF%E5%81%87%E8%AE%BE%E6%B3%95%E6%9D%A5%E5%86%991.%E8%8B%A5%E9%93%81%E8%A2%AB%E5%AE%8C%E5%85%A8%E6%B0%A7%E5%8C%96%E6%88%90Fe3%2B+%E5%88%99%E6%A0%B9%E6%8D%AE%E5%BE%97%E5%A4%B1%E7%94%B5%E5%AD%90%E5%AE%88%E6%81%92%E6%9C%89%3A%28m%2F56%29%2A3%3D%28n%2F4%2F63%29%2A3+%E5%BE%97n)
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