cho 4,8g Mg vào dd 200g dd HCl được dd A khí B
tính C% dd A
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\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{117,6.25\%}{98}=0,3\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\LTL:\dfrac{0,2}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4dư\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\DungdịchB:MgSO_4;H_2SO_4dư\\ m_{ddsaupu}=4,8+117,6-0,2.2=122\left(g\right)\\ n_{H_2SO_4dư}=0,3-0,2=0,1\left(mol\right)\\ n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\ C\%_{H_2SO_4dư}=\dfrac{0,1.98}{122}.100=8,03\%\\ C\%_{MgSO_4}=\dfrac{0,2.120}{122}.100=19,67\% \)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\
V_{H_2}=0,1.22,4=2,24l\\
m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
a) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{117,6.25\%}{98}=0,3\left(mol\right)\)
PTHH: Mg + H2SO4 --> MgSO4 + H2
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Mg hết, H2SO4 dư
PTHH: Mg + H2SO4 --> MgSO4 + H2
0,2--->0,2------>0,2---->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mdd sau pư = 4,8 + 117,6 - 0,2.2 = 122 (g)
\(\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{0,2.120}{122}.100\%=19,67\%\\C\%_{H_2SO_4.dư}=\dfrac{\left(0,3-0,2\right).98}{122}.100\%=8,03\%\end{matrix}\right.\)
a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b) Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{MgO}+m_{ddHCl}=208\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{208}\cdot100\%\approx9,13\%\)
c) PTHH: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) \(\Rightarrow\) NaOH p/ứ hết, MgCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,2\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,1\left(mol\right)=n_{MgCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{MgCl_2\left(dư\right)}=9,5\left(g\right)\\m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddA}+m_{ddNaOH}-m_{Mg\left(OH\right)_2}=402,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{402,2}\cdot100\%\approx2,91\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{9,5}{402,2}\cdot100\%\approx2,36\%\end{matrix}\right.\)
Theo gt ta có: $n_{MgO}=0,2(mol)$
a, $MgO+2HCl\rightarrow MgCl_2+H_2O$
b, Ta có: $n_{HCl}=0,4(mol)\Rightarrow x=7,3$
Bảo toàn khối lượng ta có: $m_{ddA}=208(g)$
$\Rightarrow \%C_{MgCl_2}=9,13\%$
c, Ta có: $n_{NaOH}=0,2(mol)$
$\Rightarrow n_{Mg(OH)_2}=0,1(mol)$
Bảo toàn khối lượng ta có: $m_{ddB}=208+200-0,1.58=402,2(g)$
$\Rightarrow \%C_{MgCl_2}=2,36\%$
\(a,PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)
b, Theo PTHH : \(n_{HCl}=2n_{MgO}=2.\dfrac{m}{M}=0,4\left(mol\right)\)
\(\Rightarrow x=7,3\%\)
Theo PTHH : \(n_{MgCl2}=n_{MgO}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgCl2}=19\left(g\right)\)
Mà mdd = \(m_{MgO}+m_{ddHCl}=208\left(g\right)\)
\(\Rightarrow C\%_{MgCl2}=\dfrac{m}{m_{dd}}.100\%=9,13\%\)
c, \(PTHH:MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
....................0,1..............0,2...............0,1.............0,2.......
Ta có : \(n_{NaOH}=0,2\left(mol\right)\)
=> mdd = \(m_{MgCl2}+m_{NaOH}-m_{Mg\left(OH\right)2}=213,2g\)
- Thấy sau phản ứng dung dịch B gồm NaCl ( 0,2 mol ), MgCl2 dư ( 0,1mol )
\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=11,7g\\m_{MgCl2}=9,5g\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=5,5\%\\C\%_{MgCl2}=4,46\%\end{matrix}\right.\)
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
nMg = 4,8/24 = 0,2(mol)
mHCl = \(\dfrac{m_{dd}.C\%}{100\%}=\dfrac{36,5.20\%}{100\%}=7,3\left(g\right)\)
=> nHCl = 7,3/36,5 =0,2(mol)
Lập tỉ lệ :
\(\dfrac{n_{Mg\left(ĐB\right)}}{n_{Mg\left(PT\right)}}=\dfrac{0,2}{1}=0,2\) > \(\dfrac{n_{HCl\left(ĐB\right)}}{n_{HCl\left(PT\right)}}=\dfrac{0,2}{2}=0,1\)
=> Sau phản ứng : Mg dư , HCl hết
=> dd thu được là dd MgCl2
Theo PT => nH2 = 1/2 . nHCl = 1/2 . 0,2 = 0,1(mol)
=> mH2 = 0,1 . 2 = 0,2 (g)
Theo ĐLBTKL : mdd sau pứ = mMg + mdd HCl - mH2 =4,8+ 36,5 - 0,2 =41,1(g)
Theo PT => nMgCl2 = 1/2 . nHCl = 1/2 . 0,2 =0,1(mol)
=> mMgCl2 = 0,1 . 95 = 9,5(g)
=> C%dd thu được = 9,5/41,1 . 100% =23,11%
\(Mg+2HCl(0,2)--->MgCl_2(0,1)+H_2(0,1)\)
\(nMg=0,2(mol)\)
\(mHCl=\dfrac{20.36,5}{100}=7,3\left(g\right)\)
\(\Rightarrow nHCl=0,2\left(mol\right)\)
So sánh \(\dfrac{nMg}{1}=0,2>\dfrac{nHCl}{2}=0,1\)
=> Mg còn dư, chọn nHCl để tính
Theo PTHH: \(nMgCl_2=0,1\left(mol\right)\)
\(\Rightarrow mMgCl_2=9,5\left(g\right)\)
\(nH_2=0,1\left(mol\right)\)
\(\Rightarrow mH_2=0,2\left(g\right)\)
\(m dd sau =4,8+36,5-0,2=41,1(g)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{41,1}.100\%=23,11\%\)
`Mg + 2HCl -> MgCl_2 + H_2↑`
`0,2` `0,2` `0,2` `(mol)`
`n_[Mg] = [ 4,8 ] / 24 = 0,2 (mol)`
`-> C%_[MgCl_2] = [ 0,2 . 95 ] / [ 4,8 + 200 - 0,2 . 2 ] . 100 ~~ 9,3 %`
Mg+2HCl->Mgcl2+H2
0,2-----------------0,2
n Mg=0,2 mol
=>C%=\(\dfrac{0,2.95}{4,8+200-0,2.2}100=9,29\%\)