\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{t^0}2Fe+3H_2O\)
\(0,075\rightarrow0,225\) \(0,15\)
\(V_{H_2}=0,225.22,4=5,04\left(l\right)\)
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
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