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Có lẽ đề cho dd HCl 1M bạn nhỉ?
a, Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{HCl}=0,3.1=0,3\left(mol\right)\)
PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\), ta được HCl dư.
Theo PT: \(n_{HCl\left(pư\right)}=2n_{MgO}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,1\left(mol\right)\)
⇒ mHCl (dư) = 0,1.36,5 = 3,65 (g)
Pư không tạo H2 bạn nhé.
b, \(NaOH+HCl\rightarrow NaCl+H_2O\)
\(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2+2NaCl\)
Theo PT: \(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
\(\Rightarrow n_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\)
\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)
Phương trình hóa học :
BaCl2 + H2SO4 -----> BaSO4 + 2HCl
Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)
c) Khối lượng kết tủa :
\(m_{BaSO_4}=0,12.233=27,96\) (g)
Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ;
\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\)
c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)
\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)
d) NaOH + HCl ---> NaCl + H2O
0,24 <-- 0,24
mol mol
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
0,06 mol <-- 0,03 mol
\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)
\(V_{NaOH}=0,3.2=0,6\left(l\right)\)
a. PTHH: 3NaOH + AlCl3 ---> Al(OH)3↓ + 3NaCl (1)
Ta có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{100}.100\%=12\%\)
=> mNaOH = 12(g)
=> \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
Ta lại có: \(C_{\%_{AlCl_3}}=\dfrac{m_{AlCl_3}}{200}.100\%=13,35\%\)
=> \(m_{AlCl_3}=26,7\left(g\right)\)
=> \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)
Ta thấy: \(\dfrac{0,3}{3}< \dfrac{0,2}{1}\)
Vậy AlCl3 dư
Theo PT(1): \(n_{Al\left(OH\right)_3}=\dfrac{1}{3}.n_{NaOH}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
=> \(m_{Al\left(OH\right)_3}=0,1.78=7,8\left(g\right)\)
b. Ta có: \(m_{dd_{NaCl}}=12+200-7,8=204,2\left(g\right)\)
Theo PT(1): \(n_{NaCl}=n_{NaOH}=0,3\left(mol\right)\)
=> \(m_{NaCl}=0,3.58,5=17,55\left(g\right)\)
=> \(C_{\%_{NaCl}}=\dfrac{17,55}{204,2}.100\%=8,59\%\)
c. PTHH: 2Al(OH)3 ---to---> Al2O3 + 3H2O (2)
Theo PT(2): \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> \(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
\(a)n_{BaCl_2}=\dfrac{240}{1,12}:1000\cdot1=\dfrac{3}{14}mol\\ n_{H_2SO_4}=\dfrac{122.20}{100}:98=\dfrac{61}{245}mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{3:14}{1}< \dfrac{61:245}{1}\Rightarrow H_2SO_4.dư\\ n_{BaSO_4}=n_{BaCl_2}=n_{H_2SO_4}=\dfrac{3}{14}mol\\ m_{kt}=m_{BaSO_4}=\dfrac{3}{14}\cdot233=50g\\ c)C_{\%H_2SO_4\left(dư\right)}=\dfrac{\left(61:245-3:14\right)98}{240+122-50}\cdot100=1,2\%\)
\(n_{FeCl_3}=\dfrac{325.5}{162,5.100}=0,1mol\)
FeCl3+3KOH\(\rightarrow\)Fe(OH)3\(\downarrow\)+3KCl
\(n_{KOH}=3n_{FeCl_3}=3.0,1=0,3mol\)
\(m_{KOH}=0,3.56=16,8gam\)
C%KOH=\(\dfrac{16,8}{112}.100=15\%\)
\(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,1mol\)
\(m_{Fe\left(OH\right)_3}=0,1.107=10,7gam\)
\(n_{KCl}=3n_{FeCl_3}=0,3mol\rightarrow m_{KCl}=0,3.74,5=22,35gam\)
\(m_{dd}=325+112-10,7=426,3gam\)
C%KCl=\(\dfrac{22,35}{426,3}.100\approx5,24\%\)
a)
$CuCl_2 + 2NaOH \to Cu(OH)_2 + 2NaCl$
$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$
b)
$n_{CuCl_2} = 0,01(mol) ; n_{NaOH} = 0,01(mol)$
Ta thấy :
$n_{CuCl_2} : 1 > n_{NaOH} : 2$ nên $CuCl_2$ dư
$n_{CuO} = n_{Cu(OH)_2} = \dfrac{1}{2}n_{NaOH} = 0,005(mol)$
$m_{CuO} = 0,005.80 = 0,4(gam)$
c) $V_{dd} = 0,04 + 0,06 = 0,1(lít)$
$n_{CuCl_2\ dư} = 0,01 - 0,005 = 0,005(mol)$
$n_{NaCl} = n_{NaOH} = 0,01(mol)$
$C_{M_{CuCl_2}} = \dfrac{0,005}{0,1} = 0,05M$
$C_{M_{NaCl}} = \dfrac{0,01}{0,1} = 0,1M$
nFeCl3=\(\text{325.5162,5.100=0,1mol}\)
\(\text{FeCl3+3KOH→Fe(OH)3↓+3KCl}\)
nKOH=3nFeCl3=3.0,1=0,3mol
mKOH=0,3.56=16,8gam
C%KOH=16,8112.100=15%
nFe(OH)3=nFeCl3=0,1mol
mFe(OH)3=\(\text{0,1.107=10,7gam}\)
nKCl=3nFeCl3=0,3mol
→mKCl=0,3.74,5=22,35gam
mdd=325+112−10,7=426,3gam
C%KCl=\(\text{22,35426,3.100≈5,24%}\)
thankssss!!!!!!!!!!!!
