Bài 6:
PTHH: \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\uparrow\)
\(Ba\left(OH\right)_2+FeSO_4\rightarrow BaSO_4\downarrow+Fe\left(OH\right)_2\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Ba}=\dfrac{54,8}{137}=0,4\left(mol\right)=n_{Ba\left(OH\right)_2}\\n_{FeSO_4}=0,3\cdot1=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Ba(OH)2 còn dư
\(\Rightarrow n_{Fe\left(OH\right)_2}=0,3\left(mol\right)=n_{BaSO_4}\)
\(\Rightarrow m_{rắn}=0,3\cdot90+0,3\cdot233=96,9\left(g\right)\)
Bài 5:
PTHH: \(Mn+FeSO_4\rightarrow MnSO_4+Fe\)
Ta có: \(\left\{{}\begin{matrix}n_{Mn}=\dfrac{5,5}{55}=0,1\left(mol\right)\\n_{FeSO_4}=0,3\cdot0,5=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) FeSO4 dư
\(\Rightarrow\left\{{}\begin{matrix}n_{MnSO_4}=0,1\left(mol\right)\\n_{FeSO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MnSO_4}=0,1\cdot151=15,1\left(g\right)\\m_{FeSO_4}=0,05\cdot152=7,6\left(g\right)\end{matrix}\right.\)
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