100分求一道物理题,If aluminum (Al),with an atomic weight of 27,combines with oxygen (O),with an atomic weight 16,to form the compound aluminum oxide (Al2O3),how much oxygen would be required to react completely with 52g of aluminum

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100分求一道物理题,If aluminum (Al),with an atomic weight of 27,combines with oxygen (O),with an atomic weight 16,to form the compound aluminum oxide (Al2O3),how much oxygen would be required to react completely with 52g of aluminum
100分求一道物理题,
If aluminum (Al),with an atomic weight of 27,combines with oxygen (O),with an atomic weight 16,to form the compound aluminum oxide (Al2O3),how much oxygen would be required to react completely with 52g of aluminum

翻译：如果原子量为27的铝（Al ）与原子量为16的氧（O）结合,形成复合氧化铝（Al2O3 ）,将需要多少氧气与铝52克完全反应?

4Al + 3O2¬ = 2Al2O3

4*27 3*2*16 2(27*2+16*3)

108 96 204

52g x

所以：108/52=96/x,得x=46.22g

所以需要的氧气质量为46.22克

擦~明明是道化学题好不好？不要欺负我不认识英文！
2AI+3O2=2AI2O3
AI物质的量=52/27 (mol)
所需 O2物质的量=(52/27)*(3/2)=26/9 (mol)
所需氧气的质量=26/9*32=92.444444444444444444444444克

问的是要多少氧气才能完全反应52g的铝，方程式：4Al+3O2=2Al2O3，需要氧气量=52*6*16/4*27=46.2g

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