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\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a, \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{0,6.36,5}{500}.100\%=4,38\%\)
b, \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
PT: \(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)
______0,2_______0,6______________0,2 (mol)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
\(m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)
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, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)
\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
Gọi: \(\left\{{}\begin{matrix}n_{FeCl_3}=x\left(mol\right)\\n_{MgCl_2}=y\left(mol\right)\end{matrix}\right.\)
PT: \(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_{3\downarrow}+3KCl\)
______x_________3x_________x (mol)
\(MgCl_2+2KOH\rightarrow Mg\left(OH\right)_{2\downarrow}+2KCl\)
____y_________2y_________y (mol)
Ta có: \(n_{KOH}=0,2.2,5=0,5\left(mol\right)\)
⇒ 3x + 2y = 0,5 (1)
m kết tủa = 16,5 ⇒ 107x + 58y = 16,5 (2)
Từ (1) và (2) ⇒ x = y = 0,1 (mol)
\(\Rightarrow C_{M_{FeCl_3}}=C_{M_{MgCl_2}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
\(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_3\downarrow+3KCl\\ MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\)
\(n_{KOH}=0,2\cdot2,5=0,5\left(mol\right)\)
Đặt nFeCl₃ trong 500ml X là a mol, nMgCl₂ trong 500ml X là b mol
\(\Rightarrow\left\{{}\begin{matrix}3a+2b=0,5\\107a+58b=16,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow C_MFeCl_3=\dfrac{0,1}{0,5}=0,2\left(M\right)\\ C_MMgCl_2=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
a, \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, \(n_{MgCl_2}=0,2.0,25=0,05\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,05.40=2\left(g\right)\)
c, \(n_{NaOH}=2n_{MgCl_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{0,1.40}{15\%}=\dfrac{80}{3}\left(g\right)\)
a) \(n_{NaOH}=\dfrac{60.11,2\%}{40}=0,168\left(mol\right)\)
PTHH: \(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
0,168---->0,084----->0,084
b) \(m_{kt}=m_{Mg\left(OH\right)_2}=0,084.58=4,872\left(g\right)\)
c) \(C\%_{MgCl_2}=\dfrac{0,084.95}{190}.100\%=4,2\%\)
mHCl= 3,65%.400= 14,6(g) => nHCl=14,6/36,5=0,4(mol)
a) PTHH: Mg +2 HCl -> MgCl2 + H2
0,2__________0,4_____0,2____0,2(mol)
V(H2,đktc)=0,2.22,4=4,48(l)
b)mMg=0,2.24=4,8(g)
c) mMgCl2= 0,2.95=19(g)
mddMgCl2= 400+4,8 - 0,2.2= 404,4(g)
=> C%ddMgCl2= (19/404,4).100=4,698%
\(n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\)
a. \(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)
0,1 0,1 0,1 0,2
b. \(m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\)
c. \(C\%_{Ba\left(OH\right)_2}=\dfrac{0,1.171.100}{200}=8,55\%\)
d. \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\)
0,1 0,2
=> \(a=m_{dd.HCl}=\dfrac{0,2.36,5.100}{30}=\dfrac{73}{3}\left(g\right)\)
a) \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)
b) \(n_{CaCl_2}=0,2.1=0,2\left(mol\right)\)
=> \(n_{AgCl}=2n_{CaCl2}=0,4\left(mol\right)\)
=> \(m_{AgCl}=0,4.143,5=57,4\left(g\right)\)
c) \(n_{AgNO_3}=2n_{CaCl2}=0,4\left(mol\right)\)
=> \(CM_{AgNO_3}=\dfrac{0,4}{0,4}=1M\)
\(m_{KOH}=\dfrac{200\cdot5,6}{100}=11,2\left(g\right)\)
\(\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2mol\)
\(2KOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\)
0,2 0,1 0,1 0,2
a)\(m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\)
\(C\%=\dfrac{9,5}{400}\cdot100\%=2,375\%\)
b)\(m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\)