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a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,2.1,35=0,27\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,27}{3}\), ta được Al dư.
Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,27\left(mol\right)\Rightarrow V_{H_2}=0,27.22,4=6,048\left(l\right)\)
b, \(n_{Al\left(pư\right)}=\dfrac{2}{3}n_{H_2SO_4}=0,18\left(mol\right)\)
\(\Rightarrow m_{Al\left(pư\right)}=0,18.27=4,86\left(g\right)\)
c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,09\left(mol\right)\)
\(\Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,09}{0,2}=0,45\left(M\right)\)
\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
\(0.1.............0.2.................0.1..........0.1\)
\(C_{M_{Na_2SO_4}}=\dfrac{0.1}{0.3+0.2}=0.2\left(M\right)\)
\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)
\(0.1.............0.1\)
\(m_{CuO}=0.1\cdot80=8\left(g\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=0,015\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,015}{0,1}=0,15\)
=> Tạo muối NaHCO3
\(CO_2\left(0,015\right)+NaOH\left(0,015\right)\rightarrow NaHCO_3\left(0,015\right)\)
\(\Rightarrow C_{M_{NaHCO_3}}=\dfrac{0,015}{0,015}=1\left(M\right)\)
Ta có: \(C\%_{NaOH}=\dfrac{8}{8+100}.100\%\approx7,41\%\)
\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
V dd sau hòa tan = 100 (ml) = 0,1 (l)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
Theo de bai ta co
So mol cua chat tan Na can dung la
nNa=\(\dfrac{2,3}{23}=0,1mol\)
\(\Rightarrow\) CM=\(\dfrac{n}{Vdd}=\dfrac{0,1}{0,2}=0,5M\)
a) Fe + 2HCl $\to$ FeCl2 + H2
n Fe = 5,6/56 = 0,1(mol) ; n HCl = 0,2.2 = 0,4(mol)
Ta có :
n Fe / 1 = 0,1 < n HCl / 2 = 0,2 nên HCl dư
b)
n HCl pư = 2n Fe = 0,2(mol) => n HCl dư = 0,4 - 0,2 = 0,2 mol
n FeCl2 = n Fe = 0,1 mol
Vậy :
CM FeCl2 = 0,1/0,2 = 0,5M
CM HCl dư = 0,2/0,2 = 1M
$n_{NaOH} = 2.1 = 2(mol)$
Sau khi pha, $V_{dd} = \dfrac{2}{0,1} = 20(lít)$
Suy ra :
$V_{H_2O} = 20 - 2 = 18(lít) = 18 000(ml)$
mà $D_{H_2O} = 1(g/ml)$
$\Rightarrow m_{H_2O} = D.V = 18 000(gam)$
$n_{H_2O} = \dfrac{18 000}{18} = 1000(mol)$
\(\left\{{}\begin{matrix}n_{NaOH\left(dd.1M\right)}=0,3\left(mol\right)\\n_{NaOH\left(dd.1,5M\right)}=0,2.1,5=0,3\left(mol\right)\end{matrix}\right.\)
\(n_{NaOH\left(dd.sau\right)}=n_{NaOH\left(dd.1M\right)}+n_{NaOH\left(dd.1,5M\right)}=0,3+0,3=0,6\left(mol\right)\)
\(V_{dd\left(sau\right)}=300+200=500\left(ml\right)=0,5\left(l\right)\)
\(\Rightarrow CM_{dd\left(sau\right)}=\frac{0,6}{0,6}=1,2M\)
\(\left\{{}\begin{matrix}m_{dd.sau}=500.1,05=525\left(g\right)\\m_{NaOH}=06.40=24\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Dd\left(spu\right)}=\frac{24}{525}.100\%=4,57\%\)
nHCl=0,6 mol
FeO+2HCl-->FeCl2+ H2O
x mol x mol
Fe2O3+6HCl-->2FeCl3+3H2O
x mol 2x mol
72x+160x=11,6 =>x=0,05 mol
A/ CFeCl2=0,05/0,3=1/6 M
CFeCl3=0,1/0,3=1/3 M
CHCl du=(0,6-0,4)/0,3=2/3 M
B/
NaOH+ HCl-->NaCl+H2O
0,2 0,2
2NaOH+FeCl2-->2NaCl+Fe(OH)2
0,1 0,05
3NaOH+FeCl3-->3NaCl+Fe(OH)3
0,3 0,1
nNaOH=0,6
CNaOH=0,6/1,5=0,4M
\(n_{NaOH}=\dfrac{4}{40}=0.1\left(mol\right)\)
\(n_{NaOH}=0.2\cdot1=0.2\left(mol\right)\)
\(C_{M_{NaOH}}=\dfrac{0.1+0.2}{0.2}=1.5\left(M\right)\)