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a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Ta có: \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)
b, \(m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)
c, \(m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)
a) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
b) Ta có: \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)=n_{FeSO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,01\cdot152=1,52\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=0,01\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01\cdot98}{19,6\%}=5\left(g\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)
e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
Khối lượng muối FeSO 4 tạo thành là : 0,01 x 152 = 1,52 (gam).
Thể tích khí hiđro sinh ra : 0,01 x 22,4 = 0,224 (lít).
Ta có: \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)
\(PTHH:H_2SO_4+Fe--->FeSO_4+H_2\)
a. Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(V_{H_2}=0,01.22,4=0,224\left(lít\right)\)
b. Ta có: \(m_{H_2SO_4}=0,01.98=0,98\left(g\right)\)
Ta lại có: \(C_{\%_{H_2SO_4}}=\dfrac{0,98}{m_{dd_{H_2SO_4}}}.100\%=19,6\%\)
\(\Rightarrow m_{dd_{H_2SO_4}}=5\left(g\right)\)
Ta có: \(n_{Fe}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
a. PTHH: Fe + 2HCl ---> FeCl2 + H2
b. Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,04\left(mol\right)\)
=> \(m_{FeCl_2}=0,04.127=5,08\left(g\right)\)
=> \(V_{H_2}=0,04.22,4=0,896\left(lít\right)\)
c. Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,04=0,08\left(mol\right)\)
=> \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{2,92}{m_{dd_{HCl}}}.100\%=5\%\)
=> \(m_{dd_{HCl}}=58,4\left(g\right)\)
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)
4.1)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + H2SO4 (loãng) ---> ZnSO4 + H2
0,1---->0,1---------------------------->0,1
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) \(m_{ddH_2SO_4}=\dfrac{0,1.98}{19,6\%}=50\left(g\right)\)
4.2)
Gọi kim loại hóa trị II là R; \(n_{AgCl}=\dfrac{2,87}{143,5}=0,02\left(mol\right)\)
PTHH: \(RCl_2+2AgNO_3\rightarrow R\left(NO_3\right)_2+2AgCl\downarrow\)
0,01<-----------------------------------0,02
=> \(M_{RCl_2}=\dfrac{0,95}{0,01}=95\left(g/mol\right)\)
=> \(M_R=95-71=24\left(g/mol\right)\)
Mà R có hóa trị II => R là Magie (Mg)
Câu 5:
a) Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Al_2O_3}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)
=> 80a + 102b = 14,2 (1)
nHCl = 0,2.3,5 = 0,7 (mol)
PTHH:
CuO + 2HCl ---> CuCl2 + H2O
a------>2a
Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
b------->6b
b) 2a + 2b = 0,7 (2)
Từ (1), (2) => a = 0,05; b = 0,1
=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{14,2}.100\%=28,17\%\\\%m_{Al_2O_3}=100\%-28,17\%=71,83\%\end{matrix}\right.\)
a) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
b) Ta có: \(n_{Fe}=\frac{0,56}{56}=0,01\left(mol\right)=n_{FeSO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,01\cdot152=1,52\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{H_2SO_4}=n_{Fe}=0,01mol\)
\(\Rightarrow m_{H_2SO_4}=0,01\cdot98=0,98\left(g\right)\) \(\Rightarrow m_{ddH_2SO_4}=\frac{0,98}{19,6\%}=5\left(g\right)\)