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\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(2A+2nH_2O\rightarrow2A\left(OH\right)_n+nH_2\)
\(\dfrac{0.1}{n}........................0.05\)
\(M_A=\dfrac{3.9}{\dfrac{0.1}{n}}=39n\)
Với : \(n=1\rightarrow A=39\)
\(A:K\)
\(m_{KOH}=0.1\cdot56=5.6\left(g\right)\)
\(m_{ddX}=3.9+46.2-0.05\cdot2=50\left(g\right)\)
\(C\%_{KOH}=\dfrac{5.6}{50}\cdot100\%=11.2\%\)
\(b.\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0.1....................0.2\)
\(m_{KOH}=0.2\cdot56=11.2\left(g\right)\)
\(m_{dd_X}=\dfrac{11.2}{28\%\%}=40\left(g\right)\)
$n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$2M + 2H_2O \to 2MOH + H_2$
Theo PTHH :
$n_M = 2n_{H_2} = 0,1.2 = 0,2(mol)$
$\Rightarrow M_M = \dfrac{4,6}{0,2} = 23(Natri)$
Ta có :
$m_{H_2O} = D.V = 1.200 = 200(gam)$
Sau phản ứng :
$m_{dung\ dịch} = m_M + m_{H_2O} - m_{H_2} = 4,6 + 200 - 0,1.2 = 204,4(gam)$
$C\%_{NaOH} = \dfrac{0,2.40}{204,4}.100\% = 3,91\%$
Đáp án B
bài 1: gọi công thức oxit: AO
PTHH: AO+2HCl=>ACl2+H2
\(\frac{8}{A+16}\): \(\frac{13}{A+35,5.2}\)
ta có pt: \(\frac{8}{A+16}=\frac{13}{A+71}\)<=>71A+71.8=13A+16.13
=> A
*Phản ứng vừa đủ
Ta có: \(n_{H_2}=\dfrac{3,024}{22,4}=0,135\left(mol\right)\) \(\Rightarrow m_{H_2}=0,135\cdot2=0,27\left(g\right)\)
Bảo toàn khối lượng: \(m_{HCl}=m_{muối}+m_{H_2}-m_{KL}=42,36\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{42,36}{36,5}=\dfrac{2118}{1825}\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{\dfrac{2118}{1825}}{0,5}\approx2,32\left(M\right)\)
Gọi số mol H2 sinh ra là a (mol)
=> nHCl = 2a (mol)
Theo ĐLBTKL: mkim loại + mHCl = mmuối + mH2
=> 17,5 + 36,5.2a = 31,7 + 2a
=> a = 0,2 (mol)
=> V = 0,2.22,4 = 4,48 (l)
mCl-=mA-mKL=14,2g⇒nCl-=0,4⇒nH2=0,2(mol)⇒V=0,2.22,4=4,48(l)
`1)`
`n_{Al}={2,7}/{27}=0,1(mol)`
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`
`0,1->0,15->0,05->0,15(mol)`
`V_{dd\ H_2SO_4}={0,15}/1=0,15(l)=150(ml)`
`->V=150`
`V'=V_{H_2}=0,15.22,4=3,36(l)`
`C_{M\ X}=C_{M\ Al_2(SO_4)_3}={0,05}/{0,15}=1/3M`
`2)`
`n_{Fe}={2,8}/{56}=0,05(mol)`
`Fe+2HCl->FeCl_2+H_2`
`0,05->0,1->0,05->0,05(mol)`
`V_{dd\ HCl}={0,1}/1=0,1(l)=100(ml)`
`->V=100`
`V_{H_2}=0,05.22,4=1,12(l)`
`C_{M\ FeCl_2}={0,05}/{0,1}=0,5M`
Ta có:
\(\text{a,nH2SO4=0,05.2=0,1mol}\)
\(\text{2ROH+H2SO4=R2SO4+2H2O}\)
=>nROH=0,2mol
2R+2H2O=2ROH+H2
=>nR=0,2mol
=>MR=4,6/0,2=23 => R là Na\(\text{b, C%A=0,2.40.100:(4,6+150-0,1.2)=5,17%}\)
\(\text{mdd A=4,6+150-0,1.2=154,8g}\)
=>1/2dd A là 154,8/2=77,4g
=>nNaOH=77,4.5,17%:40=0,1mol
\(\text{nCuSO4=100.12%:160=0,075mol}\)
\(\text{CuSO4+2NaOH=Cu(OH)2+Na2SO4}\)
=> Spu dư 0,025mol CuSO4, tạo ra 0,05mol Cu(OH)2 kết tủa, 0,05mol Na2SO4
Dd B gồm Na2SO4 và CuSO4 có mdd=77,4+100-0,05.98=172,5g
\(\text{C%CuSO4=0,025.160.100:172,5=2,32%}\)
\(\text{C%Na2SO4=0,05.142.100:172,5=4,12% }\)
- Thấy Cu không phản ứng với HCl .
\(\Rightarrow m_{cr}=m_{Cu}=6,4\left(g\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
.x.......................................1,5x.........
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
.y....................................y.............
Theo bài ra ta có hệ : \(\left\{{}\begin{matrix}27x+56y+6,4=17,4\\1,5x+y=0,4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\) ( mol )
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=5,4\\m_{Fe}=5,6\end{matrix}\right.\) ( g )
b, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
.......0,1.........0,2...............................
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
...0,2.......0,6..........................
\(\Rightarrow n_{NaOH}=0,2+0,6=0,8< 1\)
=> Trong B còn có HCl dư .
\(NaOH+HCl\rightarrow NaCl+H_2O\)
...0,2..........0,2....................
=> Dư 0,2 mol HCl .
\(\Rightarrow n_{HCl}=2n_{H_2}+0,2=1\left(mol\right)\)
\(\Rightarrow m_{ddB}=17,4+250-6,4-0,8=260,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{260,2}.100\%\approx2,8\%\\C\%_{FeCl_2}\approx4,88\%\\C\%_{AlCl_3}\approx10,26\%\end{matrix}\right.\)
Vậy ....
2R+2 H2O--->2ROH+H2
n\(_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)=>m\(_{H2}=0,2\left(g\right)\)
Theo pthh
n\(_R=2n_{H2}=0,2\left(mol\right)\)
M\(_R=\frac{4,6}{0,2}=23\)
=> R là Na
Theo pthh
n\(_{NaOH}=2n_{H2}=0,2\left(mol\right)\)
C% NaOH=\(\frac{0,2.40}{200+4,6-0,2}.100\%=3,91\%\)
NaOH +HCl---->NaCl +H2O
Theo pthh
n\(_{HCl}=n_{NaOH}=0,2\left(mol\right)\)
V\(_{HCl}=\frac{0,2}{2}=0,1\left(l\right)=100ml\)
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