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Td với H2SO4:
\(n_{H_2}=\dfrac{2,9748}{24,79}=0,12mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{Al}=\dfrac{0,12.2}{3}=0,08mol\)
Td với HNO3:
\(n_{Al}=a=0,08mol\\ n_{Cu}=b\)
Khí hoá nâu trong không khí → NO
\(n_{NO}=\dfrac{3,664}{24,79}=0,16mol\\ 3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\\ Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)
\(\Rightarrow a+\dfrac{2}{3}b=0,16\\ \Leftrightarrow0,08+\dfrac{2}{3}b=0,16\\ \Leftrightarrow b=0,12mol\\ \Rightarrow m=0,08.27+0,12.64=9,84g\)
a)
$Zn + 4HNO_3 \to Zn(NO_3)_2 + 2NO_2 + 2H_2O$
$Cu + 4HNO_3 \to Cu(NO_3)_2 + 2NO_2 + 2H_2O$
b)
Gọi $n_{Zn} = a(mol) ; n_{Cu} = b(mol)$
Ta có :
$65a + 64b = 3,23$
$n_{NO_2} = 2a + 2b = 0,1$
$\Rightarrow a = 0,03 ; b = 0,02$
$\%m_{Zn} = \dfrac{0,03.65}{3,23}.100\% = 60,37\%$
$\%m_{Cu} = 100\% -60,37\% = 39,63\%$
c)
$n_{HNO_3} = 2n_{NO_2} = 0,2(mol)$
$C_{M_{HNO_3}} = \dfrac{0,2}{0,1} = 2M$
$m_{Zn(NO_3)_2} = 0,03.189 = 5,67(gam)$
$m_{Cu(NO_3)_2} = 0,02.188 = 3,76(gam)$
\(\text{Đ}\text{ặt}:n_{Al}=a\left(mol\right);n_{Cu}=b\left(mol\right)\left(a,b>0\right)\\ Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\\ Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}27a+64b=7,75\\3.22,4a+2.22,4b=7,84\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Al}=\dfrac{0,05.27}{7,75}.100\approx17,419\%\\ \Rightarrow\%m_{Cu}\approx82,581\%\\ b,n_{HNO_3}=6a+4b=0,7\left(mol\right)\\ C_{M\text{dd}HNO_3}=\dfrac{0,7}{0,14}=5\left(M\right)\)
a/
\(n_{NO_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
\(Zn+4HNO_3\rightarrow Zn\left(NO_3\right)_2+NO_2+2H_2O\)
0,16 0,16 0,16 (mol)
\(\rightarrow\%m_{Zn}=\dfrac{0,16.65}{11,9}.100\approx87,4\%4\%\)
\(\Rightarrow\%m_{Fe}=100-87,4=12,6\%\)
b/
\(Zn\left(NO_3\right)_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaNO_3\)
0,16 0,16 (mol)
\(\rightarrow m_{Zn\left(OH\right)_2}=0,16.98=15,68\left(g\right)\)
1)
Zn + 2HCl --> ZnCl2 + H2
2Al + 6HCl --> 2AlCl3 + 3H2
Zn + 4HNO3 --> Zn(NO3)2 + 2NO2 + 2H2O
2)
TN2:
\(n_{NO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Zn + 4HNO3 --> Zn(NO3)2 + 2NO2 + 2H2O
_____0,05<--------------------------0,1
=> nZn = 0,05 (mol)
TN1:
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
____0,05--------------------->0,05
2Al + 6HCl --> 2AlCl3 + 3H2
0,1<-----------------------0,15
=> m = 0,05.65 + 0,1.27 = 5,95(g)
Chất rắn không tan là Vàng $\Rightarrow m_{Au} = 1,97(gam)$
Gọi $n_{Cu} = a(mol) ; n_{Ag} = b(mol) \Rightarrow 64a + 108b = 6,05 - 1,97(1)$
$n_{NO_2} = \dfrac{1,792}{22,4} = 0,08(mol)$
Bảo toàn e : $2a + b = 0,08(2)$
Từ (1)(2) suy ra : a = 0,03 ; b = 0,02
$\%m_{Au} = \dfrac{1,97}{6,05}.100\% = 32,6\%$
$\%m_{Cu} = \dfrac{0,03.64}{6,05}.100\% = 31,7\%$
$\%m_{Ag} = 100\% - 32,6\% - 31,7\% = 36,7\%$
a/
\(n_{NO_2}=\dfrac{3,581}{22,4}=0,16\left(mol\right)\)
\(n_{HCl}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)
\(Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2o\)
0,08 0,16 (mol)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,12 0,12 0,18 (mol)
\(\rightarrow\%m_{Cu}=\dfrac{0,08.64}{0,08.64+0,12.27}.100=40\%\)
\(\rightarrow\%m_{Al}=100\%-40\%=60\%\)
b/
\(n_{NaOH}=2,5.0,168=0,42\left(mol\right)\)
\(Al\left(Cl\right)_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)
bđ: 0,12 0,42 0 0 (mol)
pư: 0,12 0,36 0,12 0,36 (mol)
dư: 0 0,06 0 0 (mol)
\(\rightarrow m_{Al\left(OH\right)_3}=0,12.78=9,36\left(g\right)\)
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