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Gọi x, y lần lượt là số mol của Ca và Na2CO3.
Ca + 2HCl ----> CaCl2 + H2
x x (mol)
Na2CO3 + 2HCl ----> 2NaCl + CO2 + H2O
y y (mol)
a, Theo đề ra, ta có:
40x + 106y = 22,6
22,4x + 22,4y = 8,96
=> x = 0,3
y = 0,1
=> mCa = 0,3.40 = 12 (g)
=> %Ca = \(\dfrac{12.100\%}{22,6}\)= 53%
=> %Na2CO3 = 100 - 53 = 47%
a)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
=> 56a + 24b = 18,4 (1)
PTHH: Fe + 2HCl --> FeCl2 + H2
a-->2a------>a------>a
Mg + 2HCl --> MgCl2 + H2
b--->2b------->b------>b
=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (2)
(1)(2) => a = 0,2 (mol); b = 0,3 (mol)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{18,4}.100\%=60,87\%\\\%m_{Mg}=\dfrac{0,3.24}{18,4}.100\%=39,13\%\end{matrix}\right.\)
b) \(n_{HCl\left(pư\right)}=2a+2b=1\left(mol\right)\)
=> \(n_{HCl\left(tt\right)}=\dfrac{1.125}{100}=1,25\left(mol\right)\)
=> mHCl(tt) = 1,25.36,5 = 45,625 (g)
=> \(a=\dfrac{45,625.100}{18,25}=250\left(g\right)\)
c)
mdd sau pư = 18,4 + 250 - 0,5.2 = 267,4 (g)
\(C\%_{FeCl_2}=\dfrac{0,2.127}{267,4}.100\%=9,5\%\)
\(C\%_{MgCl_2}=\dfrac{0,3.95}{267,4}.100\%=10,66\%\)
$a)$
Đặt $n_{Al}=x(mol);n_{Fe}=y(mol)$
$\to 27x+56y=13,75(1)$
Bảo toàn e: $1,5x+y=n_{H_2}=\dfrac{11,2}{22,4}=0,5(2)$
Từ $(1)(2)\to x=0,25(mol);y=0,125(mol)$
$\to \%m_{Al}=\dfrac{0,25.27}{13,75}.100\%\approx 49,09\%$
$\to \%m_{Fe}=100-49,09=50,91\%$
$b)$
Bảo toàn H: $n_{HCl}=2n_{H_2}=1(mol)$
$\to a=\dfrac{1.36,5.120\%}{18,25\%}=240(g)$
$c)$
Bảo toàn Al,Fe: $n_{AlCl_3}=0,25(mol);n_{FeCl_2}=0,125(mol)$
$m_{dd_{HCl(p/ứ)}}=\dfrac{1.36,5}{18,25\%}=200(g)$
Ta có $m_{dd\, sau}=13,75+200-0,5.2=212,75(g)$
$\to \begin{cases} C\%_{AlCl_3}=\dfrac{0,25.133,5}{212,75}.100\%=15,69\%\\ C\%_{FeCl_2}=\dfrac{0,125.127}{212,75}.100\%=7,46\% \end{cases}$
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+65y=21,4\\x+y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4g\)
\(\Rightarrow m_{Zn}=0,2.65=13g\)
\(\%m_{Fe}=\dfrac{8,4}{21,4}.100=39,25\%\)
\(\%m_{Zn}=100\%-39,25\%=60,75\%\)
\(m_{FeCl_2}=0,15.127=19,05g\)
\(m_{ZnCl_2}=0,2.136=27,2g\)
a) 2Al + 6HCl -> 2AlCl3 + 3H2
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol
=>%mAl=20,93% =>%mAl2O3 = 79,07%
b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g
mddY=12,9+100-0,15.2=112,6g
mAlCl3=22,5g=>C%=19,98%
\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)
\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.02\)
\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)
\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)
a)
Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$
$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$
Từ (1)(2) suy ra : a = 0,1; b = 0,2
$m_{Zn} = 0,1.65 = 6,5(gam)$
$m_{Al} = 0,2.27 = 5,4(gam)$
b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$
Zn + 2HCl => ZnCl2 + H2
Na2CO3 + 2HCl=> 2NaCl + H2O + CO2
MY = 0,5875.32 = 18,8
áp dụng sơ đồ đường chéo ta đc nH2 : nCO2 = 3:2
mà nH2 = nZn ; nCO2 = nNa2CO3
=> nZn = 3/2 nCO2
ta có \(65.\frac{3}{2}x+106x=4,07\left(g\right)\) => x= 0,02 mol => nZn =0,03
a. => % na2CO3 = \(\frac{0,02.106}{4,07}.100\%=52,088\%\)
=> % Zn = 47,912%
b. nHCl pư = 2 .nZn + 2. nNa2CO3 = 2.0,03+ 2.0,02 = 0,1
=> mHCl pư = 0,1.36,5 = 3,65 (g)
=> m HCl dùng = 3,65.120% = 4,38 (g)
=> mdd HCl = \(\frac{4,38.100}{25}=17,52\)
=> mdd = 4,07 + 17,52 - 0,03.2-0,02.44 = 20,65(g)
mHCl dư = 4,38 - 3,65 = 0,73(g)
C% HCl dư = \(\frac{0,73}{20,65}.100\%\) = 3,535%