\(a)Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(1mol\) \(1mol\) \(1mol\)
\(0,1mol\) \(0,1mol\) \(0,1mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)
\(m_{H_2SO_4}=n.M=0,1.98=9,8\left(g\right)\)
\(b)Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{Fe}=n.M=0,5.56=28\left(g\right)\)
\(c)Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(1mol\) \(1mol\) \(1mol\)
\(0,2mol\) \(0,2mol\) \(0,2mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
\(\text{Ta thấy }H_2SO_4\text{ dư,Fe phản ứng hết.Bài toán tính theo Fe}\)
\(m_{FeSO_4}=n.M=0,2.152=30,4\left(g\right)\)
