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Ta có: \(n_{Fe_2O_3}=\dfrac{9,6}{160}=0,06\left(mol\right)\)
a. PTHH: Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O (1)
Theo PT(1): \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,06=0,18\left(mol\right)\)
=> \(m_{H_2SO_4}=0,18.98=17,64\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{17,64}{m_{dd_{H_2SO_4}}}.100\%=9,8\%\)
=> \(m_{dd_{H_2SO_4}}=180\left(g\right)\)
b. Ta có: \(m_{dd_{Fe_2\left(SO_4\right)_3}}=9,6+180=189,6\left(g\right)\)
Theo PT(1): \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,06\left(mol\right)\)
=> \(m_{Fe_2\left(SO_4\right)_3}=0,06.400=24\left(g\right)\)
=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}=\dfrac{24}{189,6}.100\%=12,66\%\)
c. PTHH: Fe2(SO4)3 + 3BaCl2 ---> 3BaSO4↓ + 2FeCl3 (2)
Theo PT(2): \(n_{BaSO_4}=3.n_{Fe_2\left(SO_4\right)_3}=3.0,06=0,18\left(mol\right)\)
=> \(m_{BaSO_4}=0,18.233=41,94\left(g\right)\)
Theo PT(2): \(n_{BaCl_2}=n_{BaSO_4}=0,18\left(mol\right)\)
=> \(m_{BaCl_2}=0,18.208=37,44\left(g\right)\)
Ta có: \(C_{\%_{BaCl_2}}=\dfrac{37,44}{m_{dd_{BaCl_2}}}.100\%=10,4\%\)
=> \(m_{dd_{BaCl_2}}=360\left(g\right)\)
a)
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$
$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$
b)
$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư
$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,3 0,3 0,3 (mol)
a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)
\(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)
b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
0,3 0,3
=> mCu=n.M=0,3.64=19,2(g)
Dd B chứa NaOH.
PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(n_{NaCl}=\dfrac{4,68}{58,5}=0,08\left(mol\right)\)
Theo PT: \(n_{NaOH\left(80\left(g\right)dd\right)}=n_{NaCl}=0,08\left(mol\right)\)
\(\Rightarrow n_{NaOH\left(200\left(g\right)dd\right)}=\dfrac{0,08.200}{80}=0,2\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(\left\{{}\begin{matrix}23n_{Na}+62n_{Na_2O}=5,4\\n_{Na}+2n_{Na_2O}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,1\left(mol\right)\\n_{Na_2O}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\)
Ta có: m dd B = mA + mH2O - mH2
⇒ 200 = 5,4 + mH2O - 0,05.2
⇒ mH2O = 194,7 (g)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
_____0,1----->0,15
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)
2Fe(OH)3+3H2SO4(loãng)→ Fe2(SO4)3+ 6H2O
(mol) 0,15 0,225
\(n_{Fe\left(OH\right)_3}=\dfrac{m}{M}=\dfrac{16,05}{107}=0,15\left(mol\right)\)
\(->m_{H_2SO_4}=n.M=0,225.98=22,05\left(g\right)\)
Ta có:
\(C\%=\dfrac{m_{H_2SO_4}}{m_{ddH_2SO_4}}.100\%=7,35\%\)
<=> \(m_{ddH_2SO_4}=\dfrac{22,05.100}{7,35}=300\left(g\right)\)
Chọn câu: A