Bài 1: 16 (g) Fe2O3 + HCl nồng độ 14,6%
→FeCl3 + H2O
a) Tính khối lượng dung dịch HCl đã dùng
b) Tính khối lượng FeCl3
Bài 2: Hòa tan hoàn toàn Mg vào 100 (g) dung dịch H2SO4 nồng độ 4,9% vừa đủ thu được MgSO4 Và H2
a) Tính Vh2 đktc
b) tính khối lượng Mg đã dùng
Bài 3:5,4 (g) Al + 200 (ml) dung dịch HCl→AlCl3 + H2
a) Tính VH2 = ?
b) tính MAlCl3 =?
c)tính CMHCl = ?
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\(a,n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{HCl}=0,4(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ b,n_{H_2}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ c,n_{FeCl_2}=0,2(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%\approx 22,93\%\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
a) Ta có: \(n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,05mol\) \(\Rightarrow m_{ddHCl}=\dfrac{0,05\cdot36,5}{18\%}\approx10,14\left(g\right)\)
b) Theo PTHH: \(n_{CaCl_2}=n_{CO_2}=n_{CaCO_3}=0,025\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,025\cdot111=2,775\left(g\right)\\m_{CO_2}=0,025\cdot44=1,1\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Zn}+m_{ddHCl}-m_{CO_2}=11,54\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{2,775}{11,54}\cdot100\%\approx24,05\%\)
CaCO3+2HCl→CaCl2+CO2↑ +H2O
\(+n_{CaCO_3}=\dfrac{2,5}{100}=0,025\left(mol\right)\)
\(+n_{HCl}=2n_{CaCO_3}=0,05\left(mol\right)\)
\(+m_{HCl}=0,05.98=4,9\left(gam\right)\)
\(+m_{dungdịchHCl}=\dfrac{4,9}{18}.100\%=27,2\left(gam\right)\)
\(+n_{CaCl}=n_{CaCO_3}=0,025\left(mol\right)\)
\(+m_{CaCl_2}=0,025.111=2,775\left(gam\right)\)
Theo ĐLBTKL ta có:
\(m_{CaCl_2}=2,5+27,2-0,025.44-0,025.18=28,15\left(gam\right)\)
C%=\(\dfrac{2,775}{28,15}.100\%\approx9,85\%\)
\(n_{Fe_2O_3}=0,2(mol)\\ Fe_2O_3+6HCl \to 2FeCl_3+3H_2O\\ n_{HCl}=1,2(mol)\\ V_{ddHCl}=\frac{250}{1,25}=200(ml)=0,2(l)\\ CM_{HCl}=\frac{1,2}{0,2}=6M$\)
\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)
a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)
\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a+b) Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{FeCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,6\cdot36,5}{14,6\%}=150\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{FeCl_3}=\dfrac{32,5}{150+16}\cdot100\%\approx19,58\%\)
b) PTHH: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PTHH: \(n_{KOH}=n_{HCl}=0,6\left(mol\right)\) \(\Rightarrow V_{KOH}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)
$n_{Zn}=\dfrac{6,4}{65}=\dfrac{32}{325}(mol)$
$Zn+2HCl\to ZnCl_2+H_2\uparrow$
$\to n_{HCl}=2n_{Zn}=\dfrac{64}{325}(mol)$
$\to m_{dd_{HCl}}=\dfrac{\dfrac{64}{325}.36,5}{3,65\%}\approx 196,92(g)$
Bài 1: a) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)
=> \(m_{ddHCl}=\dfrac{0,6.36,5}{14,6\%}=150\left(g\right)\)
b) \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
=> \(m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)
a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: \(n_{H_2}=n_{H_2SO_4}=\dfrac{4,9\%.100}{98}=0,05\left(mol\right)\)
=> \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
b)Theo PT: \(n_{Mg}=n_{H_2SO_4}=0,05\left(mol\right)\)
=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)