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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)
$n_{HNO_3} = 2n_{NO_2} = 0,2(mol)$
$C_{M_{HNO_3}} = \dfrac{0,2}{0,1} = 2M$
$m_{Zn(NO_3)_2} = 0,03.189 = 5,67(gam)$
$m_{Cu(NO_3)_2} = 0,02.188 = 3,76(gam)$
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)
\(\rightarrow m_{Zn\left(OH\right)_2}=0,16.98=15,68\left(g\right)\)
\(\text{Đ}\text{ặt}:n_{Al}=a\left(mol\right);n_{Cu}=b\left(mol\right)\left(a,b>0\right)\\ Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\\ Cu+4HNO_3\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}27a+64b=7,75\\3.22,4a+2.22,4b=7,84\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Al}=\dfrac{0,05.27}{7,75}.100\approx17,419\%\\ \Rightarrow\%m_{Cu}\approx82,581\%\\ b,n_{HNO_3}=6a+4b=0,7\left(mol\right)\\ C_{M\text{dd}HNO_3}=\dfrac{0,7}{0,14}=5\left(M\right)\)
,35 gam X -------------------------> m gam chất rắn khan ( Mg ; Fe ; Al ; Cl và SO4 )
nH+ = 0,5 mol
nH2 = 0,175 => nH = 0,35 => X hết ; H+ dư 0,15 mol
nSO4 = 0,125 mol
nCl = 0,25
TH1: HCl: 0,15 => nCl (muối) = 0,1
=> m = 5,35 + 0,1.35,5 + 0,125.96 = 20,9 gam
TH2: H2SO4: 0,075 => nSO4 (muối) = 0,05
=> m = 5,35 + 0,25.35,5 + 0,05.96 = 19,025 gam ( không có đáp án )
===> A
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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