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a, PTHH:
\(A+2HCl\rightarrow ACl_2+H_2\left(1\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(2\right)\)
\(AlCl_3+4NaOH\rightarrow NaAlO_2+3NaCl+2H_2O\)
b, Ta có \(n_{AlCl_3}=n_{NaAlO_2}=\dfrac{2,7}{82}=0,03\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=n_{AlCl_3}=0,03\left(mol\right)\\n_{H_2\left(2\right)}=\dfrac{3}{2}n_{AlCl_3}=0,045\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=27.0,03=0,81\left(g\right)\\n_A=n_{H_2\left(1\right)}=\dfrac{1,68}{22,4}-n_{H_2\left(2\right)}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_A=2,49-0,81=1,68\left(g\right)\\n_A=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow M_A=\dfrac{1,68}{0,03}=56\left(g/mol\right)\Rightarrow A\) là \(Fe\)
c, \(m_{\text{muối}}=m_{FeCl_2}+m_{AlCl_3}\)
\(=127.n_{Fe}+133,5.n_{Al}\)
\(=127.0,03+133,5.0,03=7,815\left(g\right)\)
C1 :
- Hòa tan hh vào dd HCl :
Mg + 2HCl => MgCl2 + H2
Fe + 2HCl => FeCl2 + H2
X : MgCl2 , FeCl2 , HCl dư
Y : Cu
Z : H2
- Dung dịch X + NaOH :
MgCl2 + 2NaOH => Mg(OH)2 + 2NaCl
FeCl2 + 2NaOH => Fe(OH)2 + 2NaCl
Kết tủa T : Mg(OH)2 , Fe(OH)2
- Nung T :
Mg(OH)2 -to-> MgO + H2O
4Fe(OH)2 + O2 -to-> 2Fe2O3 + 4H2O
Chất rắn : MgO , Fe2O3
C2:
Đặt : nCl2 = x (mol) , nO2 = y (mol)
nA = x + y = 0.6 (mol) (1)
mCl2 + mO2 = 48.15 - 19.2 = 28.95 (g)
=> 71x + 32y = 28.95 (2)
(1),(2) :
x = 0.25 , y = 0.35
Đặt : nMg = a (mol) , nAl = b (mol)
Mg => Mg+2 + 2e
Al => Al+3 + 3e
Cl2 + 2e => 2Cl-1
O2 + 4e => 2O2-
BT e :
2a + 3b = 0.25*2 + 0.35*4 = 1.9
mB = 24a + 27b = 19.2
=> a = 0.35
b = 0.4
%Mg = 0.35*24/19.2 * 100% = 43.75%
\(\text{nNa = 2,3:23= 0,1 mol}\)
\(\text{nBa = 13,7:137 = 0,1 mol}\)
\(\text{2 Na + 2H2O → 2NaOH + H2 ↑}\)
\(\text{Ba + 2H2O → Ba(OH)2 + H2 ↑}\)
\(\text{Cm (NaOH) = 0,1 : 0,2 = 0,05M}\)
\(\text{Cm(Ba(OH)2 = 0,1 : 0,2 = 0,05M}\)
\(\text{b) nCuSO4 = 0,5 .1,5 = 0,75 mol}\)
\(\text{2NaOH + CuSO4 → Na2SO4 + Cu(OH)2 ↓}\)
\(\text{ 0,1 → 0,05 → 0,05 → 0,05}\)
\(\text{Ba(OH)2 + CuSO4 → BaSO4 ↓ + Cu(OH)2 ↓}\)
\(\text{0,1 → 0,1 → 0,1 → 0,1}\)
\(\text{mY = mBaSO4 + mCu(OH)2 = 0,1.233+ (0,1+0,05).98=38g}\)
\(\text{nCuSO4 dư = 0,75 - 0,05 - 0,1 = 0,6mol}\)
\(\text{mZ = mCuSO4 dư + mNa2SO4}\)
\(\text{= 0,6.160+0,05.142=103,1g}\)
a)2Na+2H2O---->2NaOH+H2
Ba+2H2O-->Ba(OH)2+H2
n\(_{Na}=0,1\)==> n\(_{NaOH}=0,1\left(mol\right)\)
C\(_M\left(NaOH\right)=\frac{0,1}{0,2}=0,5\left(M\right)\)
n\(_{Ba}=0,1\left(mol\right)\Rightarrow n_{Ba\left(OH\right)2}=0,1\left(mol\right)\)
C\(_{M\left(Ba\left(OH\right)2\right)}=\frac{0,1}{0,2}=0,5\left(M\right)\)
b)2 NaOH+CuSO4---->Cu(OH)2+Na2SO4
0,1------------0,05------------0,05--------0,05
Ba(OH)2+CuSO4---->BaSO4+Cu(OH)2
0,1----------0,1------------0,1-----------0,1
m\(_{BaSO4}=0,1.233=23,3\left(g\right)\)
m\(_Y=14,7+23,3=38\left(g\right)\)
C\(_MNa2SO4=\frac{0,05}{0,5}=0,1\left(M\right)\)
