Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
a. PTHH: Fe + 2HCl ---> FeCl2 + H2
Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\)
=> \(V_{H_2}=0,15.22,4=3,36\left(lít\right)\)
b. Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,15=0,3\left(mol\right)\)
=> \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{10,95}{100}.100\%=10,95\%\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
\(V=0,15\cdot22,4=3,36\left(l\right)\)
\(m_{FeCl_2}=0,15\cdot127=19,05\left(g\right)\)
\(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)
BTKL:\(m_{ddsau}=8,4+100-0,3=108,1\left(g\right)\)
\(C\%=\dfrac{19,05}{108,1}\cdot100\%=17,62\%\)
Tiếp câu b:
Ta có: \(m_{dd_{FeCl_2}}=100+8,4-0,15.2=108,1\left(g\right)\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\)
=> \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
=> \(C_{\%_{FeCl_2}}=\dfrac{19,05}{108,1}.100\%=17,62\%\)
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