Bài 1
3KOH + FeCl3-----> 3KCl +Fe(OH)3
Ta có
m\(_{KOH}=\frac{400.8,4}{100}=33,6\left(g\right)\)
n\(_{KOH}=\frac{33,6}{56}=0.6\left(mol\right)\)
Theo pthh
n\(_{Fe\left(OH\right)3}=\frac{1}{3}n_{KOH}=0,2\left(mol\right)\)
m\(_{Fe\left(OH\right)3}=0,2.107=21,4\left(g\right)\)
Bài 2
2Na +2H2O-----> 2NaOH + H2
Ta có
n\(_{Na}=\frac{2,3}{23}=0,1\left(mol\right)\)
Theo pthh
n\(_{H2}=\frac{1}{2}n_{Na}=0,05\Rightarrow m_{H2}=0,1\left(g\right)\)
Theo pthh\
n\(_{NaOH}=n_{Na}=0,1\left(mol\right)\)
C%=\(\frac{0,1.40}{2,3+97,8-0,1}.100\%=4\%\)
Chúc bạn học tốt
Bài 1
MddKOH = (400*8,4)/100 = 33,6 g
nKOH = 33,6/56 = 0,6 mol
=> nFe(OH)3 = 0,6/3 = 0,2 mol
=>mKT = mFe(OH)3 = 0,2.107 = 21,4 g