\(n_{FeCl_3}=0,15\left(mol\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+2NaCl\)
0,15.............0,45...............0,15
\(V=\dfrac{0,45}{1}=0,45\left(lít\right)\)
\(m=0,15.107=16,05\left(g\right)\)
\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
0,15...........................0,075
\(a=0,075.160=12\left(g\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,075.........0,225...................0,075
\(m_{ddH_2SO_4}=\dfrac{0,225.98}{20\%}=110,25\left(g\right)\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,075.400}{12+110,25}.100=24,54\%\)
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