Bài 8: Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
a. PTHH: \(4Fe+3O_2\overset{t^o.cao}{--->}2Fe_2O_3\left(1\right)\)
Theo PT(1): \(n_{O_2}=\dfrac{3}{4}.n_{Fe}=\dfrac{3}{4}.0,3=0,225\left(mol\right)\)
=> \(V_{O_2}=0,225.22,4=5,04\left(lít\right)\)
b. Theo PT(1): \(n_{Fe_2O_3}=\dfrac{1}{2}.n_{Fe}=\dfrac{1}{2}.0,3=0.15\left(mol\right)\)
PTHH: \(Fe_2O_3+3H_2SO_4--->Fe_2\left(SO_4\right)_3+3H_2O\left(2\right)\)
Theo PT(2): \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,15=0,45\left(mol\right)\)
=> \(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)
Bài 9 mik thấy đề hơi khó hiểu
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