Bài 3:
a)
Gọi số mol ZnO, Fe2O3 là a, b (mol)
=> 81a + 160b = 24,1 (1)
\(n_{H_2SO_4}=0,4.1=0,4\left(mol\right)\)
PTHH: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
a------->a----------->a
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
b--------->3b----------->b
=> a + 3b = 0,4 (2)
(1)(2) => a = 0,1 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%ZnO=\dfrac{0,1.81}{24,1}.100\%=33,61\%\\\%Fe_2O_3=\dfrac{0,1.160}{24,1}.100\%=66,39\%\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}C_{M\left(ZnSO_4\right)}=\dfrac{0,1}{0,4}=0,25M\\C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,4}=0,25M\end{matrix}\right.\)
\(m_{dd.H_2SO_4}=400.1,2=480\left(g\right)\)
\(m_{dd.sau.pư}=480+24,1=504,1\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,1.161}{504,1}.100\%=3,2\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{504,1}.100\%=7,935\%\end{matrix}\right.\)
c)
\(n_{NaOH}=0,85.1=0,85\left(mol\right)\)
PTHH: \(ZnSO_4+2NaOH\rightarrow Zn\left(OH\right)_2\downarrow+Na_2SO_4\)
0,1-------->0,2---------->0,1
\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)
0,1--------->0,6----------->0,2
\(Zn\left(OH\right)_2+2NaOH\rightarrow Na_2ZnO_2+2H_2O\)
0,025<-------0,05
Kết tủa gồm \(\left\{{}\begin{matrix}Zn\left(OH\right)_2:0,1-0,025=0,075\left(mol\right)\\Fe\left(OH\right)_3:0,2\left(mol\right)\end{matrix}\right.\)
=> mkết tủa = 0,075.99 + 0,2.107 = 28,825 (g)
d)
- Nếu nung đến 100%:
PTHH: \(Zn\left(OH\right)_2\underrightarrow{t^o}ZnO+H_2O\)
0,075----->0,075
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
0,2------->0,1
=> mrắn = 0,075.81 + 0,1.160 = 22,075 (g)
Vậy khi nung đến 90% thì mT = 22,075.90% = 19,8675 (g)
