Fe+2CH3COOH->(CH3COO)2Fe+H2
33\112-33\56---------------33\112
n Fe=\(\dfrac{33}{112}\) mol
=>m (CH3COO)2Fe=\(\dfrac{33}{112}\).174=51,267g
=>VCH3COOH=\(\dfrac{\dfrac{33}{56}}{3}=0,196l\)
\(n_{Fe}=\dfrac{16,5}{56}=0,29mol\)
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
0,29 0,58 0,29 0,29 ( mol )
\(V_{H_2}=0,29.22,4=6,496l\)
\(m_{\left(CH_3COO\right)_2Fe}=0,29.174=50,46g\)
\(C_{M_{CH_3COOH}}=\dfrac{0,29}{0,3}=0,96M\)