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1、常温下，钠与氧气反应生成氧化钠，化学方程式为：4Na+O2=2Na2O。 2、点燃条件下，钠与氧气反应生成过氧化钠，化学方程式为：2Na+O2=Na2O2。 当给钠加热或者点燃钠的时候，钠与氧气的反应会更剧烈，会生成过氧化钠（Na₂O₂）。 对应的化学反应方程式为：2Na + O_2 {或点燃} {===}Na_2O_2在这个反应中，我们能看到钠剧烈燃烧，发出黄色火焰，最后生成淡黄色的过氧化钠固体。 钠在空气中加热，会先熔化后燃烧，发出黄色火焰，生成一种淡黄色固体——过氧化钠。 方程式为：2na + o2 =加热= na2o2。 （钠在空气中燃烧也生成过氧化钠，不过反应条件要写“点燃”）；钠常温下在空气中会发生缓慢氧化，生成氧化钠：4na + o2 = 2na2o。
在氧气中点燃：2Na+O₂=点燃=Na₂O₂ 钠与氧气反应现象：钠迅速熔化成小球，剧烈燃烧，发出黄色火焰，有少量黄色烟，生成淡黄色固体。 ### 钠和氧气点燃的化学方程式 在化学实验中，当金属钠（Na）与氧气（O₂）在点燃的条件下反应时，会生成一种白色的固体化合物——过氧化钠（Na₂O₂）。 加热或点燃时：$2Na+O_2 =Na_2O_2 (淡黄色固体)$ 2、钠和氧气反应现象：钠在空气中点燃时，迅速熔化为一个闪亮的小球，产生黄色火焰
理论上来说应该会有超氧化钠的生成,不过高中应该不会要求吧? 在空气中放置一段时间的钠（未完全变质）在点燃时涉及以下化学方程式：
钠和氧气反应现象：钠在空气中点燃时，迅速熔化为一个闪亮的小球，产生黄色火焰。 钠受热后，与氧气剧烈反应，发出黄色火焰生成淡黄色固体 过氧化钠，化学方程式为2Na+O₂点燃===Na₂O₂. 当氧气过量时，钠与氧气反应生成氧化钠，化学方程式为 Na + O2 → NaO2。 在氧气不足的情况下，钠与氧气反应生成亚氧化钠，化学方程式为 4Na + O2 → 2Na2O。 常温下就反应生成氧化钠（Na2O ）,点燃条件下生成过氧化钠（Na2O2 ）.条件是点燃,因为钠与氧气反应是钠在氧气中燃烧,所以是点燃不是加热.
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